#geometry-and-trigonometry

1 messages · Page 317 of 1

earnest echo
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Post the original question

somber coyoteBOT
upper karma
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this is the original question

the question is $cosec^2=3cot-1$

somber coyoteBOT
mighty wharf
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like snap it

earnest echo
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Can you post it as it was given to you

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Screenshot or pic

upper karma
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it's a

earnest echo
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I suggest you turn cosec(theta) into cot(theta) then solve the Quadratic

upper karma
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uhm

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how can i do that tho?

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csc = 1/sin
cot=cos/sin

earnest echo
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1+cot²(x)=cosec²(x)

upper karma
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oh

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ok so final answer is $cot^2-3cot+2=0$

somber coyoteBOT
upper karma
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tbh idk how to get the roots

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so i'll leave it at that :)

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@upper karma you sure you aren't interested?

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We can help you through it

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can it be done algebraically?

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i mean if u can, then please help me learn

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but i wont wait for 30 more mins to ping the helpers :/

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@upper karma

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yeah so simply substitute t:=cot(x)

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and you'll have a quadratic in t

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which you can throw the quadratic formula at it

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so $t^2-3t+2=0$?

earnest echo
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How did the minus before 3t become plus

somber coyoteBOT
upper karma
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my bad sorry

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and also it is worthy first to indicate restrictions because the domain of cot isn't all reals

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i got t=2 and t=1

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how do i identify the restrictions?

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if we want to be careful

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or you could check later if those solutions work

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the answers

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what did i do wrong?

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i got t=1 and 2

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nothing yet

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t isn't the same as x

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you need to sub back

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and solve for x

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do you understand what i'm saying?

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so will it be

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cot(2)^2-3cot(2)+2

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?

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i did it on my calc and i got 736

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no

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you need to sub back

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cuz cot(2)^2 is equal to 700+

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wdumm

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if you got t=1 and t=2, you need to remember that t isn't x, but instead t=cot(x)

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so you have to solve
$\begin{cases}
\cot(x)=1 \ \cot(x)=2 \end{cases}$

somber coyoteBOT
upper karma
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do you understand why?

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yes

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alright

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then simply solve those eqns to get x

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...

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but

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i dont get how im supposed to get x

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you can either draw a triangle to help you get the angle

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yea... im skipping

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....

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ngl none of this makes sense

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then

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can i do it algebraically

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why don't ask instead of saying "yes i understand" you tell me your doubt when i'm asking you?

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bruh

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i get why we did this

so you have to solve
$\begin{cases}
\cot(x)=1 \ \cot(x)=2 \end{cases}$
@upper karma

somber coyoteBOT
upper karma
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but i dont get how im actually going to get x

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ok we figured out that t is 1 and 2 using the quadratic equation

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what next?

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i can't help your demotivation against non-algebraic process

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it's not demotivation, it's the lack of knowledge

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you can either draw a triangle to help you get the angle

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so then

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i'm not prepared to do it, cuz i cant

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i'm here to show you the way

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alright, let's do it using the triangle way

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okay

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let me draw a diagram

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alright

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well first remind you that $\cot(x)=\frac{1}{\tan(x)}=\frac{\text{adjacent}}{\text{opposite}}$

somber coyoteBOT
upper karma
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oh

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that's what idk

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but i can find a pic online that sums it up

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that's what idk
do you know what tan(x) is

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@upper karma yes i do

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wait do you not understand why it is adj/opp

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just making sure

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i do

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i just dont know how to convert those new trig functions to adj/opp

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okay

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by new trig functions i mean the cosecant, cot, and secant

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okay now we are talking

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so

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$\cot(x)=\frac{1}{\tan(x)}=\frac{1}{\frac{\sin(x)}{\cos(x)}}=\frac{\cos(x)}{\sin(x)}=\frac{\frac{\text{adjacent}}{\text{hypothenuse}}}{\frac{\text{opposite}}{\text{hypothenuse}}}=\frac{\text{adjacent}}{\text{opposite}}$

somber coyoteBOT
upper karma
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i see

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let me know if you are confused with how the skycraper fractions got simplified

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no i get it

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what's next

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we got cot(x) values

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yeah so drawing triangle

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$\cot(x)=\frac{1}{1}$

somber coyoteBOT
upper karma
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it's our first case

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we have 1/1, which if you remember cot is adj/opp

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yes

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so the triangle we want to draw has 2 legs of length 1

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let me draw the triangle

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so a and b will both be equal to 1

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it'll take like 5 minutes so you may want to take a rest or make sure you understand everything we've said so far

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so a and b will both be equal to 1
yes

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alright

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im sorry if im bothering u

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i did all parts btw, a,b,c and d

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i just need to know how to find the roots

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not bothering at all

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take a 4 min rest if you want

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while i draw it

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\begin{tikzpicture}
\draw (0,0) -- (3,3);
\draw (3,3) -- node[above=10pt, right=5pt]{$1$} (3,0);
\draw (3,0) -- node[left=10pt, below=5pt]{$1$} (0,0);
\node at (0.7,0.3) {x};
\end{tikzpicture}

somber coyoteBOT
upper karma
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@upper karma

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not a cool drawing but it suffices

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oke im back sorry for being late

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all good

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ok so i get why the x is there, and i get why we have the 1s

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and for the second part we can change them to 2s

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can you recognise the angle x of this isosceles triangle

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and for the second part we can change them to 2s
almost

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yes

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for the second one it's 2/1 not 2/2

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45, 45, 90

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exactly

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so x=45 degrees is one solution

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alright

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wait

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can you try with the 2/1

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it's another special triangle

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\begin{tikzpicture}
\draw (0,0) -- (3,3);
\draw (3,3) -- node[above=10pt, right=5pt]{$1$} (3,0);
\draw (3,0) -- node[left=10pt, below=5pt]{$2$} (0,0);
\node at (0.7,0.3) {x};
\end{tikzpicture}

somber coyoteBOT
upper karma
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30,60,90

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yes

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wait is it...?

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ok what next dude

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we didnt finish the first solution, did we?

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we did

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cuz the answer isnt 45

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what is it

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that's a

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and that's radians

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not degrees

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we're doing a

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,w 45 degrees to radians

somber coyoteBOT
upper karma
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0.785

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everything we did was correct

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the book can be incorrect btw

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i took you from here cot^2(x)-3cot(x)+2=0

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that's fine

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did i typo'd

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wait

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this is the IBDP Math AA HL book, it's brand new, and we already discovered loads of mistakes

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yeah that seems like a book's mistake

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wolfram gives me 45 degrees as a solution too

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ok i just emailed my teacher

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let's do the second solution

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and i'll leave it to him

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okay hold up a sec

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ok

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i think i confused one of your questions with another answer

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dude

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we spent 45 mins on one question

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the second sol isn't an special triangle, i was saying yes to the diagram

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yea i know that

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??

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is the rest correct tho?

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we spent 45 mins on one question
so what

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yes it is

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oh that's fine then

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ok so it's not a special triangle

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what is it then?

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either way, you shouldn't point out like that, humans make mistakes too

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yup

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what is it then?
it is an ugly solution which you can get an approx by your calc

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or by the graph

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,w arccot(2)

somber coyoteBOT
upper karma
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...

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what's arccot?

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ohfwefef

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my baddd

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lol

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dw

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,w plot cot(x)

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but can't we arccot (1)?

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instead of doing the special triangle thing?

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you can if you want an approximate answer

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how do i arccot on my calc?

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cot is 1/tan

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cot^-1 is ??

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just tan?

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but i hope you agree with me that saying x=pi/4 or x=45 degrees isn't cleaner than 0.022222222222222222222222222222222...

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lmaooo

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it defo is

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okay

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wait why is it 0.0222222?

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as i recall, i think we got 0.754342

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yea 0.7853

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did u arccot 0.7853?

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i typed a random number ngl

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bruh

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lmaoooo

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that's fine lol

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but...

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ok so solution no.2 is correct

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but why is 1 incorrect?

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and how do u arccot on a scientific calculator?

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i don't think there is an specific thing for arccot

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wait a second

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,w arccot(1)

wind shard
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just do arctan(1/x(

somber coyoteBOT
upper karma
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we need it to be 3.605...

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why is it wronggggg

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??

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we established that the book was wrong

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i mean yea

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but one part of the solution is correct

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so that's questionable

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well, wolfram agrees with me

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so it's very likely your book has that solution wrong

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well

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thank u v much

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i'll proceed with the other questions

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i already did them

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and put them into quadratics

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and then arcXXX them

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yw, good luck with the rest!

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we are here if you have any more doubts.

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yup

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thank uuu

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<3

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,w arccos(-2)

somber coyoteBOT
upper karma
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wait what

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,w arccos(1/3)

somber coyoteBOT
earnest echo
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There is no solution in real values for arccos(-2)

upper karma
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welp

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neither of them is correct :/

upper karma
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@upper karma is there another problem you need help on

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nope im heading off to sleep

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none of the questions worked

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i did the arcXXX to them and none worked

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if u want, i'll drop the questions and where i reached

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Better suggestion: if you want i can help you tomorrow

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nah it's fine

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i'll submit it tomorrow

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afternoon

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in 14 hours

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Okay well

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yup

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gn

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thanks for ur help

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Gn

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really means a lot

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Yw

feral plank
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Hey guys sorry to burst in here but basically, im in a pickle, I suck at math, I try and I just suck. Im doing a math test atm, and I got like maybe 6 or so questions I need help with if y'all wouldnt mind

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last one

upper karma
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IF someone can help me with geometry please dm me

silent plank
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@feral plank we are not allowed to help on tests. requesting help on tests may get you banned. #❓how-to-get-help

feral plank
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😱

upper karma
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@upper karma just post the question here

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Yeah huh only 7 q of a test

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Not 1 but 7

silent plank
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why dm

upper karma
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Good question

agile socket
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I am wondering about how the highlight applies to how you graph?

📎 image0.png
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So far I have this

agile socket
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I think got it is this ok?

vocal fog
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Just a quick question, for a right triangle, are the 2 slopes congruent or opposite reciprocals?

silent plank
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wdym by opposite reciprocal?

vocal fog
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/when finding the slopes of right triangles are they consecutive sides?

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like for perpendicular line slopes the slopes are opposite reciprocals as in 2 and -1/2

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for example, triangle “ABC” is a right triangle. would the two slopes of triangle abc be congruent or opposite reciprocals?

silent plank
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if you're referring to the legs, negative reciprocal would be better
(assuming they aren't horizontal /vertical)

vocal fog
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so the hypotenuse (the longest side), opposite side, adjacent side is what a right triangle is composed of which means that the slopes are opposite reciprocals

hollow raven
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huh?

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u mean perpendicular?

calm umbra
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Yes the numerical value of the slopes would be opposite reciprocals because that is a property of slopes of perpendicular lines

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Usually triangles are just 3 segments floating who knows where so you may not usually think of them having a slope as you would for a line that you graph on xy axes

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@vocal fog

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This would be of course for the sides that join to make the right angle

grim cairn
umbral snow
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That's one of the answers. There is another

grim cairn
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yeah how do I find it

umbral snow
grim cairn
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lol

umbral snow
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So tanθ is the slope of the terminal arm. Is there another position on the unit circle that causes the same slope?

grim cairn
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5pi/6?

umbral snow
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Almost, but you're on a good track. That's the negative slope. So that would be tanθ = -1/√3

grim cairn
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4pi/3

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oh wait 5pi/3

umbral snow
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You want 7π/6 haha

grim cairn
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ooohhhhhhhhhhhhhh

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the opposite

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ok ok

umbral snow
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For tan, it's always an "add π"

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But yeah the slope perspective helps a ton

grim cairn
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coool thank u

timber tartan
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what is tangent?

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What is cotangent then? so confused

pure cape
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well tangent has some few different meanings, if you are talking about trigonometry then tangent is a function that can be represented by a ratios of sides in a right triangle

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tan(theta) = op/adj

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cotangent or cot is the reciprocal of tan

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so cot(theta) = adj/op

timber tartan
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thanks for clearing this up for me! a huge help

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And im guess that cosine and sine have the same relationship as tangent and cotangent?

pure cape
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not exactly

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sine is op/hyp

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but cosine is adj/hyp

covert rune
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Trig names barely make sense, don't worry

pure cape
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wonder why they didnt name secant cosine and cosine secant

timber tartan
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oh god whats secant?

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Hyp/op?

upper karma
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Reciprocal of cosine

timber tartan
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oh

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ok

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hyp/adj

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so is there a hyp/opposite?

upper karma
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That's reciprocal of sine

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Cosecant

timber tartan
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oh thanks

hallow crystal
west basin
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75 is 5sqrt(3)

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not 5sqrt(5)

hallow crystal
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75 is 5sqrt(3)
@west basin
Oh, right
But even then how do I get 2 locations

west basin
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its just 1 +- 5sqrt(3)

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for the x

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so 1 + 5sqrt(3) and 1-5sqrt(3)

upper karma
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Can someone tell how to do this Question
Q.) Find an acute angle whose Sine is 0.52

west basin
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can you use a calculator?

upper karma
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yes i can use a calculator

west basin
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sin^-1(.52)

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make sure your calc is in degrees

upper karma
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ohh ok

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is this ok

west basin
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yeah

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should be like 31.3 degrees

upper karma
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yes

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its giving some error

west basin
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use the button that says sin^-1

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or it might say arcsin

upper karma
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ok

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no didnt make a difference

west basin
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weird

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works on my calculator

upper karma
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which calculator are you using

west basin
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ti-84+CE

upper karma
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pressed the thing in yellow

west basin
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yeah yellow button and then that

upper karma
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still error

west basin
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did you add the ^-1?

upper karma
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yes

west basin
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dont

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asin is the same as sin^-1

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just do asin(.52)

upper karma
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ok

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yay finally

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its 31.3 ~ 32

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thx m8

cunning salmon
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saying ||FUCK|| is now banned (WARNING PRESS SPOILER AT OWN RISK)

void wind
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Hey yall! I need to calculate sin x, only thing given is tan 9/11. Any idea how can i approach to this problem? 😊

📎 image0.jpg
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I was thinking maybe i can use “trigonometric one”, if thats a thing in english

upper karma
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yes

void wind
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Should i use trigonometric one?

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I get this when i use it. Any ideas how i can simplify it from here?

📎 image0.jpg
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I shouldn’t square it right? So it’s cos^2=1^2-(cos9/4)^2 and i could just divide both side with cos so i get. Cos=1^2-(9/4)^2

upper karma
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$1 + \tan^2(x) =\sec^2 (x)$

somber coyoteBOT
void wind
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What is sec if i may ask?

upper karma
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$1 + \tan^2(x) =\frac{1}{\cos^2(x)}$

somber coyoteBOT
upper karma
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you get this by dividing both sides of the fundamental formula by cos^2

void wind
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Oh damn. It isn’t possible to use the trigonometric 1 formula? I haven’t learned this yet.

upper karma
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$\cos^2(x)+\sin^2(x)=1 \newline \frac{\cos^2(x)}{\cos^2(x)}+\frac{\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}$

somber coyoteBOT
void wind
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Aha

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That makes sense

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So basically i just shove in tan in there which is given. But how do i/ what should add to cos?

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Or do i solve cos in the equation

void wind
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<@&286206848099549185>

paper vale
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@void wind use cos(x)=+-sqrt(1-sin^2(x))

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so u can just get a simple equation , but as it could be plus or minus u need to check 2 cases

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wait why dont u just do tan^-1(7/11) and put that in sin

void wind
paper vale
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no, u can literally find out x by doing inverse tan

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then plug that into sin

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also u cant rearrange the an equation and use that in the same equation

void wind
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But i am not allowed to answer with decimals, they need exact answers

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Also im not allowed to use inverse :D, maybe I should’ve said that before.

paper vale
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yes so if u do what i said earlier then u can easily get the solutions

void wind
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I’ll give it a try, thank you vm 🙌

deep iron
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y does AM/BC=tan theta

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nvm

pastel anchor
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yo can anyone help me with this question in finding the complex roots?

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6x^2 -21x +24+=0

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i got stuck on 21+-sqrt135/2

rain holly
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divide the equation by 3 first

pastel anchor
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no im using the quadratic formula

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now im stuck at that part of the quadratic formula

paper vale
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the roots are gonna be the same if u divide by anything

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other than 0

rain holly
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@pastel anchor ya but if u simplify it then the calculations will be easier

pastel anchor
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so u want me to simplify it and then do quadratic formula?

rain holly
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yes

pastel anchor
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i gopt 7+-sqrt-15/4

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im stuck

rain holly
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thats correct

pastel anchor
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but i need the compelx roots

paper vale
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those are the complex roots

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but write sqrt15i as it is better notation

rain holly
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u can change sqrt(-15) to i*sqrt(15)

pastel anchor
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so i can just hand in 7+-sqrt15i/4

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as the answer

rain holly
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yes

pastel anchor
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i swear for some other questions the roots were like 3+1/2i

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@rain holly why do we leave it as it is and not simplify it any further

rain holly
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because u cant

pastel anchor
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how could i do it for 4+-sqrt144/16 then

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whats the difference

rain holly
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u can simplify sqrt(144) to 12

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but in 7+-sqrt15i/4 u cant do anything like that

pastel anchor
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yea but u can do sqrt15i still

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it wil just have decimals

rain holly
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ya that counts as simplified

main meadow
earnest echo
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,rccw

somber coyoteBOT
earnest echo
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What have you tried?

humble pulsar
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solve for a, then plug it into AC

pastel anchor
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heyo guys im sutck with another question like the previous one

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x^2-6x+28=0

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im stuck at 36+-sqrt-76/2

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how can i simplify it any further

humble pulsar
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are you solving over complex or real?

pastel anchor
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im trying to get the complex roots algrebraically

humble pulsar
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ok so it's not a 36, it's 6

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since quad formula is -b+-...

pastel anchor
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oh

humble pulsar
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yeah, then to simplify it you see if there's a square # you can pull out of 76

pastel anchor
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u cant

humble pulsar
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you can

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-76 = 4 * -19

pastel anchor
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ok i was thinking of a perfect square

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so like 4sqrt-19

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?

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@humble pulsar so is the answer (6+- 2 radical 19)/2

humble pulsar
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yeah, sqrt(-19) though

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so then do you see the simplification? (Also if you're solving complex numbers you need to introduce the imaginary unit)

pastel anchor
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yea i see it but i dont see where can i add the imaginary unit

humble pulsar
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you know you can split square roots by multiplication of the radicand right?

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sqrt(ab)=sqrt(a)sqrt(b)

pastel anchor
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no i never learned that

humble pulsar
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oh well it's true (try and prove it yourself if you know sqrt(a) = a^.5)

pastel anchor
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is the i suppose to behind the 2

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btw

humble pulsar
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behind the 2?

pastel anchor
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like 6+-2isqrt19

humble pulsar
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the 2 cancels with the 2 in the denominator

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as well as a 2 in the 6

pastel anchor
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oh

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so its 3+-sqrt19?

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actually can the -19 be sqrt19i

humble pulsar
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yeah it has to be sqrt(19)i

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sqrt(-19)=sqrt(19)sqrt(-1)

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and sqrt(-1) = i by definition

pastel anchor
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so theres parentheses?

humble pulsar
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no, you use parentheses in typing math to be explicit

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the 19 would be under the square root sign

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and the i, typically, to the right of the co-efficient

main meadow
silent plank
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properties of isosceles triangles

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,rotate

somber coyoteBOT
silent plank
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given those sides, x isn't equal to y

paper vale
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yea in an isosceles triangle it is the angles on the legs which are equal

humble pulsar
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equal sides => angles opposite them are equal

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based on the idea that small sides "=" small angles

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and vice versa

paper vale
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what

humble pulsar
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Law of Sines, but x would be 75 based on it being isoceles

paper vale
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lol why would u use law of sines to show that isosceles triangles have equal angles

upper karma
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you dont need trig for that lol

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nice kane

humble pulsar
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Im not saying that, but the idea of small angles -> small sides comes from law of sines

brisk holly
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can someone help with this

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just #5

storm sun
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ping me

bitter jetty
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@storm sun i wud say yes

storm sun
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why so?

bitter jetty
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cuz they share one side which shows equale length

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and its showing the 2 others are also of equal length

storm sun
#

ah alright thanks

snow dove
#

Angle C is a right angle. Find side c

sharp plume
#

A is the adjacent side?

snow dove
#

i assume so

bitter jetty
#

c

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@snow dove use cosine

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cn someone help lol

silent plank
#

@Argan#9381 by convention a is the side opposite A, etc...

bitter jetty
#

can someone help me, idk where to start

deep iron
#

is it possible to find the height using 2 angles and 1 side?

tawdry ruin
junior light
#

@tawdry ruin What have you tried/thought so far?

upper karma
#

It says "test"

sharp plume
#

👀

tawdry ruin
#

It’s a chapter review my teacher lists everything thing as tests

runic magnet
#

hey

#

mind explaining thales theorem?

#

I don't know it that much

polar wave
runic magnet
#

huh

earnest echo
#

What have you tried?

glacial dawn
#

Find the point at which the line of (-3,12) and (6,0) intercepts the y-axis

polar wave
#

ohhhhhhhh

glacial dawn
#

If u have that u have the side lengths of the unshaded triangle

polar wave
#

yeah i can calculate those with the cords i got right

glacial dawn
#

Take the area of the whole triangle- unshaded triangle

polar wave
#

using distance formula

glacial dawn
#

Uh

#

Lemme see

#

So

#

Eqn of line is y=-4/3x + 8

#

Hence intercept is (0,8)

#

Side lengths of unshaded triangle are 8 and 6

#

Hence the area of unshaded is 24

#

Area of total triangle which is shaded and unshaded together

#

Would be

#

(1/2)bh

#

Are u following @polar wave ?

polar wave
#

yes

glacial dawn
#

So then

#

U have to find point where that (-3,12) line intercepts the x axis on the left

#

Since angle is perpendicular it will be opposite reciprocal of other angle

polar wave
#

yeah that makes sense now

glacial dawn
#

U got answer?

polar wave
#

im solving

glacial dawn
#

K

polar wave
#

thanks tho

glacial dawn
#

Np

#

Lmk if u get stuck

polar wave
#

ok

glacial dawn
#

I got 126 un^2 btw @polar wave

polar wave
#

ok thx

#

lemme check

glacial dawn
#

K

#

Btw u don’t need to use distance formula

polar wave
#

I got 126 un^2 btw @polar wave
@glacial dawn same, thanks

glacial dawn
#

Np

tawdry ruin
tawdry ruin
#

Pls

tawdry ruin
#

Pls

upper karma
#

This is not a "do this for me" server.

blazing coyote
#

Hello

#

I desperately need help on this question

#

Can anyone help me, its a geometry question

#

A bit confused

orchid perch
#

1, 2, 3, are angles

#

and 4

#

@blazing coyote the middle angels are all 90 degrees

#

because angles across from each other equal each other

#

and angles next to each other that make a straight like both equal 180

#

do you get that?

silent plank
#

you can't directly assert that the angles are 90° at this point

#

go through a few triangle congruency proofs

blazing coyote
#

ok ty

rocky ether
humble pulsar
#

@rocky ether what question in specific do you need help w/?

rocky ether
#

literally the whole thing 🙂

#

i failed algebra so idek wtf im doing

humble pulsar
#

ok well question 2 is full of scaffold so i believe you can handle that lol

#

but for 1, what you want is to keep the perimeter of the rectangle at 44, but change the actual dimensions

#

so what's the general perimeter of a rectangle?

rocky ether
#

i have 0 clue

#

idek how to find it

humble pulsar
#

perimeter is add up all the side lengths

#

so P = l + l + w + w = 2l+2w

rocky ether
#

im an all a/b student lol i have an a in geometry but ive just cheated the whole way through to hold my gpa so

#

whats l and w

humble pulsar
#

length and width of the rectangle

#

so you want 44 = 2l+2w => 22=l+w => w=22-l

rocky ether
#

that looks liike eonofuihngeouigujg to me because im stupid 🙂

upper karma
#

desperate last attempt, i'm attempting to understand unit circles and figuring out the values of trigonometric functions

#

the ones where you evaluate a point unit circle for sine cos and tan

#

becase this is all slipping over my head

#

im sorry this is late also, attempting to figure this out before its too late for me

#

its getting late, any videos to recommend?

willow kiln
#

im sorry this is late also, attempting to figure this out before its too late for me
there are people in different zones here lol (though here it's nearly 1 am and I should be studying)

#

anyhow JustAsk

upper karma
#

what

#

i mean i did, i aasked yous guys

#

It’s getting too late, I need to sleep

#

See yous

willow kiln
#

i mean i did, i aasked yous guys
ah I thought you had a question regarding a problem. Not sure if I'd recommend something other than the usual, e.g. Khan Academy

upper karma
#

My problem is that I’m just confused, end of grading period and I had to use class time to catch up on other stuff and the lesson completely flew over my head so,

#

I’m understanding radians and transferring them to degrees and back, it’s manageable but

#

Also I’d show work on what exactly I’m having trouble on but my handwriting is attrocious

#

I do not know what values I’m being asked for

high geode
#

they want all values of the six trigonometric functions by the looks of it

#

sin = y, cos = x, tan = y/x csc 1/y sec 1/x cot x/y

#

-15/17 is y and 8/17 is x

upper karma
#

my thick skull is just discovering about the other three

willow kiln
#

they're just fancy notation for $\dfrac{1}{\sin(t)}$, $\dfrac{1}{\cos(t)}$, $\dfrac{1}{\tan(t)}$

somber coyoteBOT
willow kiln
#

they want all values of the six trigonometric functions by the looks of it
but yes, they're asking you this and you should use the definition of the trig functions (as ratios between sides of a triangle)

high geode
#

csc would be 1/1/18

#

17 sorry

upper karma
#

i need to get some coffee to absorb this, be back in a second

high geode
#

sec would be 1/-15/17 and cot would be -15/17 / 8/17

#

which is -15/8

#

cot that is

upper karma
#

watching some of the khan academy vids because i'm still pretty lost

upper karma
#

idk why this still isn't clicking with me

#

im just trying to figure out the placements of these integers, where this video is even pulling them from and why

#

i think that;s what im trying to figure out anyways

#

im not so sure anymore

signal swallow
#

what are you trying to do?

upper karma
#

id like to know that myself

signal swallow
#

lol

#

what is the topic?

#

or an example question?

upper karma
#

im just stuck currently, missed a lesson and now it seems like im facing a brick wall

#

question is up there a little

signal swallow
#

okay, so it gives you an angle t

#

and gives you both an x and y coordinate based on that angle

high geode
#

six trig functions

#

the formulas to find them are in the front of that chapter in the book

#

and i posted them before

upper karma
#

the theta would be easier to figure out of the coordinates made sense to me, and yeah, i searched up the 6 formulas and i'm taking note of them but, i'm starring at this image and im just drawing blanks

📎 UC2.png
signal swallow
#

what exactly are you confused on

upper karma
#

the coordinates

signal swallow
#

what do you know about sin?

#

what does it correspond to?

upper karma
#

it corresponds to y when sin t

signal swallow
#

yes

#

and what does cos correspond to?

upper karma
#

x?

signal swallow
#

yes

#

so if it gives you P(-15/17, 8/17) you have cos and sin

#

what is tan in terms of sin and cos?

upper karma
#

that would be x/y

signal swallow
#

that's cot

upper karma
#

oh

signal swallow
#

tan is y/x

#

which you can also evaluate

upper karma
#

shoot, i got them mixed up

signal swallow
#

because you have both y and x

upper karma
#

cot and tan also correlate to being unable to correspond when x =0

#

well, because it cant, there;s no 90 degrees

#

well

#

cot would be for y, not x

signal swallow
#

yes

#

do you understand the problem now?

#

and do you know the other 3 trig functions that you must get?

upper karma
#

yea, csc and sec

signal swallow
#

yup

upper karma
#

i think my dummy thick skull is getting some of this in now

high geode
#

you just gotta look at those formulas that's all

#

I can never remember them lol

upper karma
#

isnt that why they put them on calculators

high geode
#

I think I have the same book as you

upper karma
#

i dont have a book

#

online lessons,

#

plus, i skipped alg 2

#

so im extra clueless this year

narrow crag
#

um

#

do converse statements always have the same truth value?

rocky ether
#

what does this mean?

#

and how do i find a fixed perimeter

narrow crag
#

describe a pattern for what types of conditional statements have a true inverse statement and what types of conditional statements have a false inverse statement.

dark jacinth
#

i could use some help with c and d

#

i am not totally sure how vectors work in 3 dimensions

dark jacinth
#

<@&286206848099549185>

#

please?

dark jacinth
#

anyone?

odd plover
#

(c) since FG is parallel to AD D is (0,9,0) and F is (8, 0, 12), to find DF components find the difference of destination point and initial point coordinates (8-0, 0-9, 12-0)

#

(d) Two vectors A and B are perpendicular if and only if their scalar product is equal to zero. i.e Ax Bx + Ay By + Az Bz == 0

#

check for d

upper karma
#

You know books can be mistaken too

upper karma
#

@dark jacinth c) the "components" of a vector means x, y, z. If you translate the entire model so that point D is the origin, what are the coordinates of point F?
d) the dot product between the two vectors needs to be zero for them to be perpendicular

high geode
#

Okay I'm stumped on this problem. From a boat on a river below a dam, the angle of elevation to the top of of the dam is 10degrees38. If the damn is 845 feet above the level of the river, how far is the boat from the base of the dam(to the nearest foot)?

upper karma
high geode
#

we would have to use tangent to solve it wouldn't we

upper karma
#

in order to solve any triangle, you need to know 3 things: AAS, ASS, SSS, AAA

#

A - angle
S - side

high geode
#

ah but we don't know the adjacent side

upper karma
#

you have two angles and a side

#

and since you have two angles... you can find the other one, easily, since all angles from a triangle need to add up to 180 deg

#

you can use soh cah toa

#

you can also use the law of sines

high geode
#

so a sq + b sq = c sq to solve for a right?

upper karma
#

solve for x

high geode
#

i got 4556 but that's not the right answer

#

but shouldn't that be it?

upper karma
#

no

high geode
#

I tried 834.62/tan(10.38)

upper karma
#

why 834.62?

#

it says the height is 845

high geode
#

because i thought you had to use Pythagorean Theorem

#

845/tan(10.38) doesn't give me the right answer either

upper karma
#

,w 845/tan(10.38)

somber coyoteBOT
upper karma
#

that doesn't seem right...

high geode
#

the answer is 4501 but I can't seem to get it

#

I think my calculator is supposed to be in degree mode but I can't seem to get this answer

#

i get 4613

upper karma
#

that's what I get, too

#

why do you think it's 4501?

high geode
#

45001 is what the answer says

upper karma
#

two methods, same answer

high geode
#

ya that's what I get

upper karma
#

says to round up to the nearest foot

#

4501 isn't the nearest foot, lol

high geode
#

no it isn't lol

#

I'm thinking the answer on this multiple choice is wrong

upper karma
#

"10degrees38" is 10.38, right?

#

maybe the base of the dam isn't a 90 deg angle?

#

are there any pictures?

high geode
#

it says 10degrees and then has a 38' after it

#

10° 38' is what it says

upper karma
#

I think it's safe to assume 10.38

high geode
#

ya that's what I thought

#

I guess I'll just talk to my professor about it

upper karma
#

,w 845/(tan(10.38 degrees)

somber coyoteBOT
high geode
#

ya its odd

#

but thanks for the help I think our answer is right

indigo chasm
upper karma
#

Smh don't multipost

indigo chasm
#

?

#

@upper karma what u mean

upper karma
#

Don't put it in multiple channels

paper vale
#

just set those values equal to eachother

#

as equal chord length implies equal angle

upper karma
#

@indigo chasm what result did you get?

high geode
#

anybody know how to solve number 58 I'm kinda lost

humble pulsar
#

it says use a graph, so graph y=sec(x) and y=1

paper vale
#

that is kind of silly though

#

just make it cos(x)=1 and then it is easy @high geode

upper karma
#

hello

#

can someone explain me how to solve this

paper vale
#

this is just notation

#

so u should probably learn that

#

there is no thinking involved

upper karma
#

i already got an answer but i'm not sure if it's correct

#

so the answer should be 3;5

#

right?

humble pulsar
#

@upper karma what is the actual question? write out the [-3,5) intersect (3,8) as 1 interval?

upper karma
#

i guess so

paper vale
#

yea it is just (3,5)

humble pulsar
#

^ and your logic with the 2nd pic is right

paper vale
#

why are you using semi-colons btw

humble pulsar
#

that doesnt matter lol

upper karma
#

honestly

#

i have absolutely no idea

humble pulsar
#

Like using a semi colon for intervals makes sense, as it mitigates confusion between points and open intervals w/ american notation

upper karma
#

i don't really like math, i'd rather master a programming language or something else

#

but well, i still gotta do my homework

upper karma
#

The radius of a straight circular cone is 9 cm, and the height of the cone is 12 cm. Get the slant height.

#

@upper karma it rhymes with Mythagora

#

yeah so how do i get the slant height

#

@upper karma it rhymes with Mythagora

#

what do you mean

#

oh

narrow plinth
#

Hi could someone help me with this or give some hint I can't figure it out?
the lines connecting the midpoints of the opposite sides of the quadrilateral are 6 cm and 8 cm, and one diagonal is 10 cm. Calculate the area of the quadrilateral.

worn folio
upper karma
#

have you tried anything or thought on any?

worn folio
#

@upper karma My bad i had a really big brain fart then i realized i was looking for definitions

upper karma
#

rip daniel do you know what vertical angles are

worn folio
#

yea

upper karma
#

and how they're equal?

worn folio
#

they are symmetrical

upper karma
#

no they're equal not symmetrical

#

ok so given that they're equal

#

and that x + 20, and 3x - 10 are vertical angles

#

you can conclude that

#

x + 20 = 3x - 10

#

understand?

worn folio
#

yes, looks similar to algebra

upper karma
#

so given x + 20 = 3x - 10. solve for x

worn folio
#

ok 1 sec

#

x=15

upper karma
#

yes

#

for this one

#

i've circled the vertical angles

worn folio
#

so same thing

upper karma
#

yes

worn folio
#

accept keep the same colors together?

upper karma
#

the pairs of colors, are vertical angles

worn folio
#

well for your diagram

#

ok

#

got it 1 more second

#

x=28

upper karma
#

now for the red one

#

we know that since they're vertical angles

#

x = y - 2x

#

and you're also given that x = 28

#

so solve for y now

worn folio
#

y=3x?

#

would i plug in 28 for the x?

upper karma
#

yes, that's correct

#

in that order

#

are those the only problems you needed help on

worn folio
#

y=84?

upper karma
#

yea

#

x = 28, and y = 84

worn folio
#

thanks, i might need a couple more let me check

#

I think this is the last one i need help on i think i can do the others on my own

upper karma
#

well you know why it's true right

worn folio
#

they are equal?

upper karma
#

yea, if you make a line from P to U, and another from L to M. they'll be equal

#

you have to prove it now

worn folio
#

would i start with the givens?

upper karma
#

just explain why it's true

#

as if you're talking to someone who doesn't get it

worn folio
#

ok

obtuse hornet
upper karma
#

yes

worn folio
#
  1. If i bisect this segment perfectly in half you will see that pl = um
#
  1. Because they are equal i can take L and U and pair them with the farthest point and it will remain equal.
#

Is that good so far?

upper karma
#

it's okay

#

instead of saying "If i bisect this segment perfectly in half you will see that pl = um"

#

maybe say

#

if i cut this line in half pl is still equal to um

worn folio
#

ok

#

and is number 2 good?

upper karma
#

for #2 if i didn't know i would ask why it works like that

#

i also don't see for #1 what cutting the line in half does

worn folio
#

bisecting mean perfectly cut in half right?

#

cause if you bisect the line you will see that they are the exact same

upper karma
#

i think a better explanation would be

#

if you removed the line LU

#

you would see that PU and LM both are the enitre line

worn folio
#

oh, How can i come up with better reasons like that>

#

?

upper karma
#

try explaning it by removing parts of the line

worn folio
#

ok

#

thanks

#

Thank you so much for all of the Help and i hope you have a great day or night

sage sandal
#

Is there anyone willing to help me study for a trignometry test I have this Wednesday? I've been struggling a lot and according to my professor this is the hardest test out of all of his math classes. I'm not sure if this is the correct channel to be asking this, so I apologize if it isn't.

humble pulsar
#

@sage sandal do you have any questions you need help with?

sage sandal
#

as of right now I only have one question, I'd have to switch over to my pc to bring it up though.

obtuse hornet
#

Anyone able to help me on how to do this problem?

silent plank
#

what have you tried?

obtuse hornet
#

Wait

#

x is 8 right

#

Am I an idiot this is so easy lol

#

unless i understood wrong

silent plank
#

yes. x is 8

obtuse hornet
#

Ok for some reason i thought it couldnt be that easy so i confused myself lol

silent plank
#

there is way too much information given

#

getting those 2 points is up to chance

obtuse hornet
#

Yeah i was thinking the info was unnecessary so i thought there must be more to it or something

silent plank
#

unless you give all reasonable methods

#

(or do it anyway to confirm that the problem is actually consistent)

upper karma
jagged relic
upper karma
jagged relic
upper karma
#
  1. The two points E and F lie in the plane R, then the line m containing them lies in the plane R as well. Postulate: Flat Plane Postulate and If there are at least three points.....postulate: #9
#
  1. Two planes Q and R intersect, then their intersection n is a line. Postulate #11
#
  1. Two points E and F lie in plane R, therefore line m lies in plane R. Postulate #10
jagged relic
upper karma
#
  1. No, there’s only one line than contains A and B
#
  1. Yes, the two points aren’t on a single plane.
#
  1. No
jagged relic
crystal marlin
willow kiln
#

do we suppose that the curved thing is a circle arc?

crystal marlin
#

Yes it’s a circle arc

upper karma
#

it's the arc of a quarter of a circle with radius 10

#

might be easy if you draw the whole circle inside a square to visualize

crystal marlin
#

Ok

paper vale
#

no, what u need to do is work out the area of the square and subtract the area of the quarter circle @crystal marlin

#

which is obviously 1/4 pi*r^2

#

and for perimeter do 1/4 2*pi*r +10

upper karma
#

That was for them to figure out after this hint

might be easy if you draw the whole circle inside a square to visualize

upper karma
#

@crystal marlin you know, if you put 4 of those together, you end up with a circle inside a square

#

it's easy to see the radius, and to calculate the area for the circle

#

then calculate the area of the square

#

and lastly, you subtract the area of the circle from the area of the square

#

oh, and divide by 4 as a final step 🙂

#

or else you'll get the area of all 4 corners, instead of just one

quartz prism
#

I can see that you'll get sinxcos(100x) as the first term and sin(100x)cosx in the last term

#

so you get pairs of terms

#

which will give sin100x

#

wait

#

nvm

#

I got it

#

😅

#

this was the solution I saw after solving it and I thought doing the whole 2S part's redundant but wondered why they did it and wanted to see if any term didn't have a "mirror" term but I see why you'd do this now

📎 unknown.png
long vector
#

Anyone?

crystal marlin
#

@crystal marlin you know, if you put 4 of those together, you end up with a circle inside a square
@upper karma yes I figured it out alr lol, but thanks a lot for your help

humble pulsar
#

@long vector which question do you need help with?

dim gyro
#

So this is the formula for finding the distance between two pints, on the last step do i keep simplifying or does d=41 or does d=6.403

📎 image0.jpg
calm umbra
#

d = root(41)

#

Thats the end

dim gyro
#

It says to leave answer in radical form btw

calm umbra
#

You can approximate that as a decimal or leave it as root(41) depending on what they ask

dim gyro
#

Forgot to include that part

calm umbra
#

Oh then leave as root(41) in that case

dim gyro
#

👌

calm umbra
#

Leaving them in square root is the only way to get exact answers for those

#

Unless you have a perfect square

#

Decimals will only give you a less precise approximation

dim gyro
#

Thanks

coral frigate
#

Hello, I’m have trouble trying to Google this scenario.
I have a character in my video game that has a vertical velocity of (vy).
I want to calculate the horizontal velocity (vx) that is needed for the character to land (d) amount of pixels from it’s initial position.

So, I have:
Vertical velocity (vy)
Gravity (g)
Time (t)
Distance to travel (d)

I want to find:
Horizontal velocity (vx)

Thank you!

calm umbra
#

This depends on how gravity works in your game

#

The equations we use in physics problems / projectile motion problems rely on the earth's gravitational constant

leaden pike
#

I need help with math

#

lol

calm umbra
#

What do you need help with

coral frigate
#

@calm umbra yea, the gravity in my game also works as a constant. We could say 1 pixel every frame.

#

I guess I just need help figuring out the formula to use.

calm umbra
#

Ok so im gonna assume then that acceleration caused by g = -1 distance/time

Where distance is pixels time is frames

#

The problem is jumoing from initial position to a position that is d units away horizontally?

#

Is that correct

coral frigate
#

That is correct

calm umbra
#

Ok so really what we need to figure out is how long you are in the air

#

As that will determine how much time our character has to move horizontally

#

Do you have a function for velocity from jumping?

#

Vertical velocity&

umbral snow
#

Gravity is a constant velocity in your game? That can work, but normally we see constant acceleration

calm umbra
#

I think its a constant acceleration on vertical velocity

#

We are gonna do like

vy - gt

#

So if vy, initial vertical velocity from a jump is constant and acceleration by g is constant:

Current Vertical Velocity: vy - gt

Current Vertical Position: (vy-gt)*t

#

Set vertical position equal to 0 and solve for t

coral frigate
#

Yes, constant acceleration.

So, for example:

velocity.y += gravity

As for the vertical velocity function, I might not understand the question, but the vertical velocity is set to some value once the user presses a keyboard key. So the only thing that changes the vertical velocity is gravity.

calm umbra
#

One choice will be t =0 the other choice will be the length of time youre in the air for the jump

#

Once you find the non zero value for t we go to our horizontal velocity

This one im gonna assume is constant after jumping and holding right:

Horizontal velocity has no acceleration component

So horizontal distance is xy*t

#

the t that you solved for earlier will tell you how much horizontal distance you cover in one jump

#

Set d = to that xy*t and you can solve for the horizontal velocity needed to clear thay distance in one jump

#

Hopefully that helps