geometry-and-trigonometry

[wicked narwhal]

No sorry I have to go bed but People in this channel can help u 🙂

[royal dawn]

oh ok thats fine

[fresh dagger]

anyone know how to solve this?

[rough timber]

Where should the angle a go when i add these two

[oblique gazelle]

help please just dm me the answer

[short quartz]

hey

is this server dead?

[limpid onyx]

Why is $\angle OAB = \angle OB’A’$ and $\angle OBA = \angle OA’B’$?

[somber coyote]

kappa

[limpid onyx]

(A’ is A when inverted as shown in the sketch. Same follows with B.)

[jagged mist]

yo bois

i need some help

i got an 82 in my geo class and i got a geo eoc tomorrow

i dont know any geometry

what percent would i need to pass the class idc about the test

[limpid onyx]

what’s the passing grade

[jagged mist]

i think like a 35%

or somewhere around it

[limpid onyx]

how much does the grade for the eoc weigh

[jagged mist]

im not sure conpletely just going off what ive seen online

30%

[limpid onyx]

you can get a 0 and pass

[jagged mist]

i might have an 87 but i dont know bec i used edjinuity and they closed my class so im not sure of the grade

[jagged mist]

u sure?

[limpid onyx]

you’ll be at a 57 I think

[jagged mist]

i need at least a d they said

[limpid onyx]

Note that the triangles are similar

[upper karma]

can u look in #help-32

[jagged mist]

any1 else got an answer for this

[fossil dune]

Can any1 help

[jagged mist]

quirky ah viewing angle

[upper karma]

if you would take pictures that were properly taken, maybe

[fossil dune]

Ok hold on Srry

Is that good?

[upper karma]

much better

[fossil dune]

Ok

[upper karma]

have you calculated the length of AB yet?

[fossil dune]

6# that all I have for it

[upper karma]

well (2.33, 6.33) isn't part of the line

[fossil dune]

Yea

[upper karma]

btw forget this I was thinking about vectors for some reason

[fossil dune]

That why I need help I don’t understand where I went wrong

[upper karma]

your mistake is probably that you divided by 3 instead of 4

since the ratio is 1 to 3 you need 4 parts

if I am not dumb rn

[fossil dune]

Oh ok

[upper karma]

got it?

[median radish]

Doing math corrections, can anyone explain how to do 16, 17, or 18? I need it for tomorrow

[mighty bane]

4x^2 + 24x + 20

Change the minus 24x to plus 24 x

[buoyant ingot]

16th one

sin^2 β = 3/4
=> sin β = __+__sqrt(3)/2
=> β = pi/3 , 5pi/6

[dark sparrow]

=> β = pi/3 , 5pi/6
incorrect

[smoky jetty]

For the first one as the hint stated, try to analyze the relationship of angles when parallel lines are cut by a transversal, which for this case, the bases are the parallel lines, while the congruent pair of sides of each isosceles triangle are the transversals.

[smoky jetty]

Another way to look at this problem is as the triangle is cut by a line, thus forming 2 triangles, the upper base and the line are parallel to each other, while the other sides are the transversals.

[smoky jetty]

does the two arcs of the interior angles indicate an equal measure?

[peak apex]

WW1 Eraser

[limpid onyx]

Can anyone prove OA’B’=ABO

[grave ferry]

It says the revolution on the wheel is 360 the wheel is 110m high I get on at 5m how do I write this down as a sinusoidal equation

[limpid onyx]

Never mind, I’ve figured it out. Using the converse of PoP, you can find that quadrilateral ABA’B’ is cyclic and use the fact that the exterior angle of an opposite angle is equal to that angle.

[fossil dune]

Fr

[potent forge]

i want to see someone elses work

[sage sluice]

this isnt a rectangle cus the diagonals arnt congruent right?

[wooden mango]

A_1, A_2, A_3 and so on provide infinite naming options

[warped bone]

Ooo

[smoky jetty]

seems like it

[oak matrix]

need help with these

[twin locust]

Can someone explain or prove why diagonals of rhombus bisect each other? I'm confused

[dark sparrow]

is your confusion the "I don't know what the statement means" kind or "I know what the statement means, but not why it's true" kind?

[twin locust]

I now understand it! sorry to bother you.

[wicked narwhal]

Thales theorem

[red coyote]

How do I know which side length of a triangle is it's base

[little lark]

the base is the side perpendicular to the altitude you are using

[oak matrix]

waht qeustion

[wicked narwhal]

First

Do you Know it?

[hazy barn]

H

[upper karma]

heyy i am going to start coordinate geometry and trigonometry of class 11th on jee advanced level.....any suggestions??

[smoky jetty]

have fun learning the concepts!

[red coyote]

What is the altitude I'm using

What is an altitude

[little lark]

idk

an altitude is a line that comes from a vertex that forms a right angle on the base

[red coyote]

That's the just the height

Anyways

[little lark]

yeah

the base is relative to your heigh which is an altitude

[red coyote]

What is the perpendicular bisector of a line segment called?

[red coyote]

What if I don't know the height

Do I just choose any side as the base

[little lark]

you can

[red coyote]

Ok

[red coyote]

?

Is it axis of symmetry

[little lark]

perpendicular = intersect with 90 degree angle

[red coyote]

Uh

What

What is it called

[little lark]

wdym?

i didnt read your question

[red coyote]

What is the perpendicular bisector of a line segment called

[little lark]

uh i dont remember in english probably

ill google

isn't it just called a perpendicular bisector?

median?

@red coyote ?

[little lark]

what did you write?

[red coyote]

Isn't a Median the average of sets

[red coyote]

Axis of symmetry

But I'm not sure

[little lark]

i am pretty sure its called a median

probably mislead due to my shitty translating efforts from hebrew

[red coyote]

Does a median always form a 90 degree angle on the line segment it bisects with?

It has to be axis of symmetry

[little lark]

yeah

so does a perpendicular bisector

the definition of a perpendicular lines is that they intersect at 90 degrees

[red coyote]

A median doesn't necessarily have to be perpendicular

[little lark]

yeah fair probably axis of symmetry then

since that makes sense

[potent verge]

u should prove both triangle similar then as we know similar triangles sides are in ratio (just simplify and enjoy the calculation)

[potent verge]

they dont intersect they bisect( it means dividing into two equal parts)

[little lark]

ohhh ok

idk geometry in english sorry

[potent verge]

ook geometry in what language u understand ??

[little lark]

hebrew lol

[potent verge]

lol

[little lark]

ill try to just translate the shit i know to english

[potent verge]

okie full support

[little lark]

thanks owo

[errant lake]

Any help would be appreciated

[oak matrix]

noppe, idk how to solve both. i mean i tried but i couldnt

[oak matrix]

i got 6 units on the first one

but 2 is underlined so idk which one is true

[errant lake]

For question 12 i think u should use Pythagorean theorem. One side is 13-12 so 1 the other side is x(what ur looking for) and 9 is the hypotenuse

[stuck hound]

help pls

ik its ez but i do forced online school so im dumb

[timber cargo]

Interior angles are supplementary angles, so they add up to 180 degrees

So 70+50+(6x-18)=180

[stuck hound]

thank you

[pulsar garden]

hi guys, can any tell me how to draw this pls? Given a triangle ABC, let B′ and C′ , respectively, be the projections of B and C onto the median AM or its prolongation.

im having problem to draw the projections of B and C

but my draw is kinda weird

[opal fog]

Hi, can somebody help me here?

[neon heath]

what am i doing wrong here for question C

because i end up with a equation that has no real roots

so i dont know how to simplify it

and find the Y value

[lunar birch]

Have found out

[neon heath]

?

[olive bloom]

I need help with these 3 questions please

[silent plank]

what have you tried

[still trout]

Just use inverse function, you will get answer
Fir example in 1st question just do tan(inverse)(86) you will get your answer i.e. 89.33°

[olive bloom]

oh it's that easy?

thanks bro

i overlooked that lol

[still trout]

Hell yeah

[still trout]

Nvm lol

[olive bloom]

can you also help me with this one

find x and round to the nearest 10th

What I did:

Tan(20) = 9/x
9 x Tan(20) = X
= 3.27
= 3.3

But it was wrong

What did I do wrong?

(The answer is C I just need to see the working out)

[silent plank]

you're mistake is going from

Tan(20) = 9/x
to
9 x Tan(20) = X
describe what you did without using vague words like move/transpose/put

also don't use x for multiplication (especially not when the variable x is present)

[olive bloom]

So I had to isolate X

I multiplied each side by 9 to get

9 times Tan(20) = X

[olive bloom]

oh ok

[silent plank]

9/x * 9 isn't x

[olive bloom]

wdym

[silent plank]

i mean what i said

[olive bloom]

but X is the adjacent side

that's what I need to find

[silent plank]

i know

[olive bloom]

It's the same thing in algreba

8 = 9/x
9*8 = X

for example

[silent plank]

and although multiplying both sides by 9 is valid, getting from
$$\frac 9x \cdot 9$$
to $x$ is not

[somber coyote]

ℝamonov

[silent plank]

neither is what you just posted

$$\frac{\blue{9}}{\green{x}} \cdot \blue{9} \redneq \green{x}$$
are you implying
$$\frac {\blue{0}}{\green{1}} \cdot \blue{0} \wthonk \green{1}$$
?

[somber coyote]

ℝamonov

[olive bloom]

oh uh

then what do i do

im so confused

[silent plank]

multiply / divide properly

start with fixing the simpler example

[olive bloom]

so how is it supposed to be?

[silent plank]

and solve 8 = 9/x
properly

multiply/divide both sides by the same value(s), but do it properly

[olive bloom]

8 = 9/x
x9 x9

72 = X

[silent plank]

you're doing the exact same mistake as before

[olive bloom]

what is the mistake

this is how i learnt it?

[silent plank]

the exact same one that i pointed out

you learnt wrong and/or rismembered it

are you implying that
$$\frac{1000000}{1} \cdot 1000000$$
is 1?

[somber coyote]

ℝamonov

[silent plank]

because that's essentially what you're doing

[olive bloom]

definetly the second one

im stupid when it comes to math

[silent plank]

if you started with
$$8 = \frac x9$$
then multiplying both sides by 9 would get you
$$x = 72$$
because unlike $\frac 9x \cdot9$, $\frac x9 \cdot 9$ is actually $x$

[somber coyote]

ℝamonov

[silent plank]

$\frac 9x \cdot 9$ would be $\frac{81}{x}$

[somber coyote]

ℝamonov

[olive bloom]

oh wait

i see it now

X = 9/8

right

[silent plank]

yes

[silver mesa]

I need help with this L is 1,2

[silent plank]

I need help with this L is 1,2
the diagram isn't clear enough in indicating what L is supposed to be

can you show the full problem
with instructions and relevant information

[silver mesa]

Sorry I didn’t really say what the question was, the question is what is the depth if L is 1,2 Meters

[silent plank]

and as i said

the diagram isn't clear enough in indicating what L is supposed to be

[silver mesa]

I think it’s supposed to be from the highest point to the water

[silent plank]

ok

you could do something like apply sine rule for the small triangle on the right
that'll give the the value of x

and then continue with right triangle trig on the small right triangle on the left to get the total height and hence depth

[silver mesa]

So like this? Sine(40)=1,2+D/X?

[silent plank]

()

[upper karma]

Hi guys

Can smone help me with a geometry problem?

[dark sparrow]

http://dontasktoask.com

@upper karma

[upper karma]

triangle ABG is given
with A(-5,-2) while the median BM has an equation of y=x-1 and the height DG has an equation of y=-2x-2

Find the equation of the side AB

@dark sparrow any ideas

[dark sparrow]

have you made a diagram?

[upper karma]

Diagram u mean a shape?

@dark sparrow of u mean shape yes I made one

[dark sparrow]

idk what you mean by "shape"

maybe you could show it here

[upper karma]

How can I find equation of the side AB

[dark sparrow]

okay, so that is a diagram.

not quite the diagram i was expecting but a diagram nonetheless.

anyway, i'd try looking for a way to find the coordinates of B

or, wait.

AB is perpendicular to DG and passes through the known point A.

[upper karma]

@dark sparrow how does it pass through the point a

Hmmm

[dark sparrow]

"how does AB pass through A?"

is that what you're asking?

[upper karma]

Am asking the equation of the AB

But maybe I can connect

These 2?

What's your idea

[dark sparrow]

my idea is that since you know the equation of DG, you know its slope. and from that, you will know the slope of AB.

[upper karma]

Hmmmm

[dark sparrow]

you apparently were confused how the line AB passes through A

[upper karma]

Yh lol now I get it

Is it possible for u to show me in piesh of paper?

And btw ty for your time

[dark sparrow]

no, sorry. i don't want to do your homework for you.

[upper karma]

Okey so them help me understand

Yes I do know the slope of DG but how can I connect it with the equation of AB

@dark sparrow

[dark sparrow]

what do you know about the slopes of two lines that are perpendicular to each other?

[upper karma]

That their λ×λ =-1

[dark sparrow]

the product of their slopes is -1, yes.

your notation for that is bad.

now what is the slope of DG?

[upper karma]

In my country greece that's the letter we named it with sooooo

It's y=2x-2

[dark sparrow]

yeah but you cannot use the letter lambda for two different things.

also the equation of DG is not y=2x-2. you said it was y=-2x-2.

and even then i asked you only for the slope.

[upper karma]

Yes Sr I missed it

So the λ IS - 2

Sr for the nottatio

[dark sparrow]

the slope of DG is -2.

[upper karma]

Yes

[dark sparrow]

so what is the slope of AB?

[upper karma]

It's 1/2

[dark sparrow]

indeed

so now you know the slope of AB is 1/2, meaning you can write its equation as y = (1/2)x + c, where c is a constant that you don't know yet.

[upper karma]

Okey yes

[dark sparrow]

and now since you know A is (-5,-2) you can use that info to find c

[upper karma]

OHHHHHHH

Wait

Gimme a sec

@dark sparrow is it y=1/2x +1/2

At its final form?

@dark sparrow soooo

[dark sparrow]

let me check

,calc (1/2) * (-5) + 1/2

[somber coyote]

Result:

-2

[dark sparrow]

okay yeah good

[upper karma]

Niceeee

Niw

Now

The coordinates of b

How do I approach that?

[dark sparrow]

that wasn't part of the original problem, was it...?

or maybe you did not share all of it

[upper karma]

That's the second

[dark sparrow]

anyway, you know the equations of BM and now also AB

[upper karma]

Yes

[dark sparrow]

find the intersection point of those lines

it will be B

[upper karma]

@dark sparrow I found

B(3,2)

Is that correct??

[dark sparrow]

seems ok

[upper karma]

Niceeeee

Now

Coordinates of G

[dark sparrow]

triangle ABG is given
with A(-5,-2) while the median BM has an equation of y=x-1 and the height DG has an equation of y=-2x-2

just to remind ourselves of the problem and so i don't have to scroll up for it all the time...

[upper karma]

Yepppppp

[dark sparrow]

G is a bit trickier

[upper karma]

Yepppppikkk

Any ideassss

[dark sparrow]

hold on, i might have to draw something myself...

we know the equations of AB and DG and we know that they intersect at D. so we can find the coordinates of D

draw a line through M parallel to DG, and it will intersect AB at the midpoint of A and D

use this to find the equation of that unnamed line, then find its intersection with BM (whose equation is known) to get M

then find G knowing that AM = MG

maybe there is a less complicated way to do this but maybe there isn't

[upper karma]

D is

-1,0

[dark sparrow]

you also learn a lot more info along the way

i don't want to bother checking your numbers at this point

if you doubt them open desmos and put all the points and lines on the graph

[upper karma]

I don't doubt myseld

F

@dark sparrow do after D

What else can help us to find G

[dark sparrow]

from here onward

if you want you can phrase it another way

let X be the midpoint of AD. [find the coordinates of X]
MX is parallel to DG [find the equation of MX]
MX and BM intersect at M [find the coordinates of M]
M is the midpoint of AG [find G]

[upper karma]

Wait what

I didn't understand that

[dark sparrow]

which part?

[upper karma]

What x?

What is x?

Can u show me at least the shape in a paper? Only the shape

[dark sparrow]

let X be the midpoint of AD.

i am giving a name to a new point

[upper karma]

Oooo

I will tell u in an 1.30 if u are right . Just solved it

[main oracle]

Hii

How can I get help? I’m not exactly sure

[mighty hatch]

Hi, I'm also new here and not sure, but I recommend reading the instructions in #❓how-to-get-help

[rough timber]

Why is angle c supplementary with 45

I dont see it

[tribal rover]

Can someone explain this to me

[exotic yarrow]

Adjacent angles of a parallelogram are supplementary

[exotic yarrow]

1. Look up the side splitter theorem
2. Attached image - use that to make a trig equation and solve for the variable

[tribal rover]

This is what I did for question 2 but idk how to solve it

[exotic yarrow]

Take the reciprocal of both sides for each equation

[tribal rover]

What

[exotic yarrow]

Do yk what 'reciprocal' means?

[tribal rover]

No

[exotic yarrow]

The reciprocal of a non-zero number $a$ is $\frac{1}{a}$ \ \ For instance, if we had the equation $\frac{1}{n}=\frac{1}{4}$, taking the reciprocal of both sides would yield $n=4$ \ \ An important thing to note while doing this is that if there are two fractions, you cannot simply take the reciprocal of both of them individually. For example, if $\frac{1}{n}+\frac{1}{m}=10$, this does not imply $m+n=\frac{1}{10}$.

[tribal rover]

Do I divide 4.9 by sin 18

[somber coyote]

Civil Service Pigeon

[upper karma]

I need helppp

<@&286206848099549185>

dude, plsss I don't have much time, school starts tomorrow, this is due tomorrow start of class 😭 😭 😭

[sullen wedge]

|vw|=|wx| so the angles pointing at them are also equal

The answer would be 180-114 divided by 2

33

[upper karma]

OMG TYSM

can I ask a few other, too???

[sullen wedge]

Ofc

[upper karma]

ok so idk if this is right..

[sullen wedge]

I'm not quite sure what the question is but i guess it's right

[upper karma]

it asks if the triangles are congruent

[sullen wedge]

what do those letters stand for

İn the answers

[upper karma]

oh its like.. SSS is side side side, SAS is side angle side and so on..

[sullen wedge]

Ohhh wait

1 sec

[upper karma]

HL is hypotenuse length

[upper karma]

ok!

[sullen wedge]

İts sss

Sorry if I'm mistaken tho

[upper karma]

oh no it's alr

so you're not sure?

[sullen wedge]

I'd say 90 percent

[upper karma]

alr tysm!

[sullen wedge]

Any other questions

[upper karma]

in this I filled the 2 blanks but I don't get the other ones

[sullen wedge]

My English math language is terrible 🫣

Do you have any numeric questions

[upper karma]

oh alr, 1 sec..

idk if this is right.. I'm terrible at this

[sullen wedge]

1 sec

Correct

[upper karma]

the proof is the only one left, plus there's a few more but umm they have english lol

[sullen wedge]

Send them I'll give them a try

[upper karma]

I've filled them, just not sure if they are right

[sullen wedge]

Correct

[upper karma]

this?

[sullen wedge]

1 is transitive

2 is symmetric

3 is reflexive

[upper karma]

wait but isn't ab=ab reflexive

[sullen wedge]

Well I'm not quite sure

I just googled these terms as I don't know them 😅

[upper karma]

alr then.. I'll leave that for now...

[sullen wedge]

1 sec

[upper karma]

I put them in order btw.

[upper karma]

alr

can you check this?

[sullen wedge]

Correct

[upper karma]

ONLY THIS LEFT LOL

[sullen wedge]

Btw your answers are correct here I did it wrong

[upper karma]

OH ALR!

[sullen wedge]

I'll try to solve this one

[upper karma]

ok

[sullen wedge]

4 is sas

[upper karma]

ok

and 2nd?

oh wait I got it

it's def. of bisector right?

[sullen wedge]

Uhmmm

Yes

I'm checking 4 again

[upper karma]

ok

[sullen wedge]

I think it's sss

[upper karma]

gosh I'm gonna submit it... 🤞

oh hell nah, I got a 14/15..

[sullen wedge]

What question

Was wrong

[upper karma]

you were right, I didn't correct the sss

sorryyy

I'll do it rn

[weary drift]

@upper karma is this an ongoing test?

[upper karma]

nah it's an overdue hw assignment

[sullen wedge]

Was 4 wrong

[upper karma]

I would never do this on a test or quiz dw!

[upper karma]

no I guess the 2nd part where you said sss

[sullen wedge]

🥲

Srry

[upper karma]

it's alr haha

oh btw @weary drift look, I'm not lying haha

[weary drift]

@sullen wedge try not to just give out answers

think socratic method

[upper karma]

I'm really bad at math

[sullen wedge]

Yeah my bad

[upper karma]

wait which one do you think was wrong?.. 2nd part or 4th part?

[sullen wedge]

Could be both tbh

[upper karma]

but it's says 3/4... so only 1 wrong right?

[sullen wedge]

Yup

But I'm not sure wich one is wrong

[upper karma]

ok there are 9 attempts left.. I'll try both!

[sullen wedge]

Ok

[upper karma]

wait so what were the options you gave?

[sullen wedge]

sss or sas

For the 4th

[upper karma]

oh alr!

tysm!

wait I still got only 3 points out of 4 for that question..

@weary drift can you help me?

@everyone

[upper karma]

Is this trigonometry???

[prime musk]

one question...

is it an equilateral triangle?

Then CB=AB=2 root 5....

[prime musk]

Ohh so then just use the distance formula and equate

[neat steeple]

?

You mean a^2+b^2=c^2?

[magic token]

Can anyone find this angle?

[distant timber]

73 degrees

i think

[wind coral]

@distant timber i got 146, could you explain your thought process?

[full finch]

can i know how you got 146?

[wind coral]

of course, i have not done geometry in a while so i could be wrong. first, i noticed that B is equal to the arc ADC. so i would need to find out what that arc equaled. next, i knew that the total degree of a circle is 360 so i used the equation 132+82=arcABC. then i subtracted that answer from 360 and i got 146.

[wind coral]

please feel free to correct me if i made an error

[finite hollow]

People interested in geometry, please help solve the problem

[smoky jetty]

what problem

[royal drift]

https://www.dontasktoask.com

[finite hollow]

#help-34

[finite hollow]

I'm sorry

Please help me prove that if in an acute triangle ABC the height CC1 is divided at the point of intersection of the heights in a ratio of 3:1. And the point M lies on SS1 and the angle AMC=90°, then you can continue the VM to the point F, such that СF || AB and CFTС1B - a parallelogram.

[stiff tulip]

I've a question

What is the sum of the sines of the 3 angles of a (possibly flat) triangle?

[zenith vessel]

guys i need some help

i want to learn trigonometry

what are some good problem books?

[finite hollow]

Help please #help-34

[graceful quest]

cordinate geometry we be used

[wicked narwhal]

Divide by 2 I think

[wind coral]

why would we do that?

[wicked narwhal]

No (132+82)/2=107

[nocturne remnant]

Angle at circumference is half of angle at center

[wicked narwhal]

Yes ABC is a Inscribed angle

[wind coral]

ah ok

[upper karma]

know about arcs and cyclic quadrilateral

[silent swallow]

Help pls #help-1

[vital flare]

The angles of a right triangle is in arithmetic progression whose common difference is C cm. Find it's Area.

My C = 0 cm

is this even solvable?

This is bugging me how is an angle measured in centimeters

[glacial moth]

uh yeah that doesn’t really make sense

the difference of angles can’t be centimeters?

and even when you get the angles you can’t find the area since you don’t have side lengths

[vital flare]

My own visulization

I guess it could be a 30 60 90 right triangle and the C refers to the sides?

But there is still the problem of C = 0cm

Therefore the sides are equal

Although the angles are different?

[nocturne remnant]

how did you get C = 0 anyway?

[vital flare]

It's given

Well, I think this is unsolvable

Instructions say I can write "Can't solve" if unsolvable

[upper karma]

thanx

[vital flare]

The C given is from the last digit of my age and its 0

Anti-cheating measurement ig

[lament pollen]

Anyone know how to solve this?

[silent plank]

consider exponential growth formula

[dark sparrow]

60% per second? that's some ludicrous growth rate.

[lament pollen]

Fr what I was thinking as well lmao

[lament pollen]

Hmm alr, I've never used that formula before I'll give it a try

Wait im confused, how do I make the equation using the formula
y=a(1+r)

[ornate jackal]

To find the population after one week, we need to calculate how many times the population multiplies by 1.6 over the course of one week. We can do this by raising 1.6 to a power that represents the number of seconds in one week.

One week has 60 seconds per minute, 60 minutes per hour, and 24 hours per day, so there are 60 x 60 x 24 = 86400 seconds per day.

So the equation would be (1.6)^(86400*7) = (1.6)^(604800)

And the ratio of P1 to P0 is 2.7 x 10^14

[ornate jackal]

P1 = P0(1 + 0.60)^(604800) it's y=a(1+r)^t and t is time.

[dark sparrow]

And the ratio of P1 to P0 is 2.7 x 10^14
this feels lowball-y

no way raising 1.6 to the power of several hundred thousand lands you only in the 10^14 range

,calc 1.6^604800

[somber coyote]

Result:

Infinity

[dark sparrow]

well, ok, out of bounds

,w 1.6^604800

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ann#0413

[dark sparrow]

yeah thought so

[ornate jackal]

sorry for the calculation mistakes

[smoky vector]

I need help quick on why this is wrong I don’t get it

[regal raptor]

wouldn’t it be AB?

[gusty umbra]

AB

[gusty umbra]

but if your going off the fact that ACB = 64, then it could also be DB as well due to the congruent angles

[regal raptor]

How do you come up with AB

[timber halo]

can anyone here help with a word probelem i have ?

[upper karma]

send plz

[timber halo]

ok

[upper karma]

yes? nice photo, i like the shadow. please tell us what you need help with.

[timber halo]

/

[upper karma]

which one?

[timber halo]

number 22

[upper karma]

the one which asks to multiply the polynomials?

[timber halo]

yes

[upper karma]

k

distirbute/ use FOIL

i will give answers

[timber halo]

thanks

[upper karma]

ok

yeah i am typing

A) 6x^2 - x - 2
B) 6x^3 - x^2 - 12x - 5
C) -3xy + 3y^2 + 8x - 8y
D) x^2 - 9y^2

thre u go

@timber halo

[timber halo]

thank you so much

[gusty umbra]

Since it's opposite from the largest angle

whose opposite side is not shared between the two triangles

[regal raptor]

Lmao that was litteraly what I was thinking

I second guess myself too much

[weak rapids]

I need help finding a few of the measures.
(The measure of angle AWX and XBY)
If you need any extra information let me know, I have most of the other measurements complete

[upper karma]

use the properties of kites and trapezoids

[granite mortar]

@upper karma don't just share answers.

[urban echo]

soneone help me with congruence plz

what is an angle bisector

[smoky jetty]

a segment/line that divides an angle into two equal measures?

[lucid jasper]

if a^4 + b^4 = c^4 prove 2((sinC)^2)=tanA*tanB

[prime musk]

angle bisector

[upper karma]

can someone help

[quaint rain]

Set up your proportions

They share one of the same angles so it is similar

Your proportions are 20/15=32/x

@upper karma

Then subtract x by 15

[quaint rain]

I'm interested by this but I have never learnt trig equations but know all trig functions and geometry-wise. I will look into this later, very intriguing...

[upper karma]

thank you

i appreciate the help

[loud jasper]

Can someone help

[upper karma]

How would i set this problem up

[quaint basalt]

Find the area of the semi circle and the isosceles triangle. The diameter of the circle is the base of the triangle. And you can find the area of the semi circle by getting the radius and getting the area of the circle and dividing that by 2.

[upper karma]

use proportions

oh yeah, somebody already helped

🙂

yeah

[south condor]

would it not just be proportional so 4/5 = 6/x ?

[upper karma]

yeah

[upper karma]

i trier that and got 2.75

not the right answer

[arctic basalt]

4x=30=>x=30/4=7.5

[south condor]

youd get 7.5 your crosssss multiplying wrong

it becomes 4x=30 after you do it

x=7.5

[unborn bison]

I found the slopes but I’m confused on part ii can anyone help

[south condor]

idk what they mean by opposite vertex but for perpendicular would be y=mx+b -> y= -1/m + b

[lucid jasper]

I got the solution, someone dropped it elsewhere, it was nuts, I cant believe they put something like this unironically, I was like "theres no way they want me to use this method..." And yet you had to used that method

[upper karma]

Anyone want to help me to solve this question?

A circle is drawn through the points (4; 2), (2; -3) and (9; -3). Find the equation of this circle. Tangents are drawn through these points. Find the coordinates of the points of intersection of these tangents.

[upper karma]

yea i was adding instead of multiplying

thanks

[quaint rain]

Could you possible dm me that? I want to learn it. If possible I mean, thx.

[south condor]

Area of of triangle, circles and rectangles

You just like separate them into simpler shapes

Then add their area

Triangle area equal base* .5 * height

Rectangle is base times height

And circle is A=πr^2

So a semi circle youd just divide by 2

[pulsar stream]

<@&286206848099549185>

[haughty hollow]

Where are u stuck??

[pulsar stream]

The

what is the center of diluation

and

what is the scale factor of the

diluation

[haughty hollow]

Alr

The simple bit

Is the scale factor

[pulsar stream]

Kk

[haughty hollow]

Pick any point

[pulsar stream]

Ok

[haughty hollow]

For example A

Find its cords

[pulsar stream]

Kk

[haughty hollow]

And then A'

So A is 9,3

And A' is 3,1

And u divide

9÷3=3

3÷1=3

So the factor is 3

[pulsar stream]

OK THANK I

Awhwat ab

the dilation

[haughty hollow]

Hold on I need a diagram

[pulsar stream]

Kk

[haughty hollow]

Here

Check this out

[pulsar stream]

I. Abt see it

Cant

Can u say theandwer @haughty hollow

[haughty hollow]

The pint where they all connect is the center of dilation

Point*

A line through A and A'

Another through B and B'

And a third through C and C'

@pulsar stream

[pulsar stream]

OMG THANK U

[haughty hollow]

I'm glad I could help

That Cord is the center of dilation

[pulsar stream]

OK THANKS SM

[haughty hollow]

Np

[pulsar stream]

Wait so what’s the number

[haughty hollow]

It's not a number

Its a co-ordinate

On the graph

It's where they all meet

Which is (0,0) in this case

[pulsar stream]

OHHH

[haughty hollow]

Yeah

[pallid verge]

Anyone have some good tips for remembering triangle similarity with proportions? Not full in depth things just stuff I can remember for a test

[alpine delta]

find the slop of one slid of the triangle and then take its negative reciprocal andthen createh equation y = slope(x-x1)+y1 simplifie and you have the equation of a line that is perpedniculare to that segment of the triangle and intersects the opposite vertex

[lethal holly]

all three angles are the same = their corresponding sides are in proportion

[sinful remnant]

Any angle is coterminal to itself, right?

[silent plank]

what's your issue with these questions

[dense elk]

If you're stuck with the solution use, sin theta = perpendicular/hypotenuse, cos theta = base/ hypotenuse , sin theta = cos(90-theta).

Remember the perpendicular and base varies according to the angle your considering

For example if you take angle X, YZ will be perpendicular and XZ will be the base

[pliant pumice]

Hello! Polygon ABCDE is irregular and complex, but is it convex? All internal angles are less than 180, but I am not sure...

[nocturne remnant]

The word “convex” is only used for polygons which are “simple” - ones which don’t self-intersect. So no

[pliant pumice]

Ok, thank you!

[solemn nebula]

thank you

[errant lake]

Any help please

[errant lake]

Idk how to draw this

[patent chasm]

Can someone tell me how to find a number diameter when you only know the radius?

[wind coral]

isnt the diameter of a circle twice the radius?

[wind coral]

just multiply the radius of the circle by 2 then youll have the diameter

[patent chasm]

Oh that much easier than I thought. LOL

[wind coral]

ah ok, good luck on whatever your doing 🙂

[patent chasm]

What about this?

[wind coral]

should be 110

[patent chasm]

Can you help me get the answer, not tell me the answer plz.

[wind coral]

ahh ok

[patent chasm]

Sorry I’m just trying to learn.

[wind coral]

so first since the sides are at a ratio of 6 to 5

hold on let me send a pic

nvm i cant find my phone

but simply i just did this
6 to 5
12 to 10
18 to 15
24 to 20
30 to 25

i scaled it up untill i came across a number that can be multiplied to get 750 sq in

in this case that was 30 * 25

so 30 and 25 will be the length of the sides of the rectangle

now to find the perimeter i just added 30+25

and multiplied by 2

[wind coral]

its all good :p

do you have any questions about how i did this?

[patent chasm]

Nope makes sense. Thanks now I just have one last question to do.

[wind coral]

np, good luck

[patent chasm]

Can you help me with the last one too?

[wind coral]

sure ill see if i can, send a pic

[patent chasm]

Number 13.

[wind coral]

first of all it cant be a because if n=2 then it would not equal 4.75

cant be c either because if n=1 that equation would equal 4.75, which doesnt match the table

cant be e because of n=1 then that function would be equal to 0

which doesnt match the table as well

and d is just clearly wrong lol

which only leaves b as the answer

@patent chasm did ya understand well?

[patent chasm]

I figured it out. Thanks.

[wind coral]

np 🙂

[timber cargo]

Represent the side lengths in terms of a variable, let's say x. You know that the ratio between the side lengths is 6:5 or 6x:5x. And the product of those two terms equals to 750. Since you know that, you can solve for x and find the perimeter

[wind coral]

lol he explained how the ratio part worked better than me :p

thx

[timber cargo]

👍

[rapid sigil]

How do I solve for theta? $$0 = \frac{sin^2\theta}{2}$$

[somber coyote]

I'mNemo

[wise pawn]

can you solve for sin(theta) first

[somber coyote]

I'mNemo

[rapid sigil]

The starting equation is this

[wise pawn]

Uh oh

[buoyant ingot]

can someone confirm the answer for me?

[runic beacon]

What answer did you get?

[buoyant ingot]

38

[tulip vine]

you cant solve this $\sin^2 \theta\in[0,1]$ but here $\sin^2 \theta=2$ which is not possible

[somber coyote]

Miso Soup

[nocturne remnant]

are you sure lol

[tulip vine]

yes?

[nocturne remnant]

Oh wait you’re talking about the other pic

Idk

Oh they made the mistake

Soz

[nocturne remnant]

Maybe ||u = 0||

[sleek verge]

How to solve for x in radians?

[upper karma]

is this from physics wallah?

[buoyant ingot]

idk someone shared it

I just solved it

it is maybe from there

[upper karma]

the font and color coding of the question looked like from there so yeah.

nice to know

[buoyant ingot]

:)

[graceful quest]

0

[pine nacelle]

Is this correct? My answers are in red

[upper karma]

Yes, all of your answers are correct/

[upper karma]

If the condition is that all angles belong to the subset of wholes numbers, then 15 degrees and 25 degrees are the only answer. If the condition is that all angles can belong to subset of rational number (excluding the negative numbers and there can be no negative angles or etc in 2-d geometry or even 3-d geometry until you reach upper dimensions), then 15, 22, and 25 can be answers. In no case what so ever can 35 be an answer and that would result in angle K being 60 degrees thus automatically making the other two angles being set to 60 degrees. This violates the question's diagram as only angle L and angle J can be congruent to each other due to the rule that states that the bases of isosceles triangle are congruent.

[true radish]

could someone help me understand the question in G? How would I solve for this I know that the x intercepts are whatever degrees would have a y value of 0 like 90 or 270 but how would i demonstrate that with any of these options?

[neon girder]

anyone can show me how to solve it??

its about linear velocity

[worldly niche]

see

if you want the car at 11 miles per hour, the cable should also move at 11 miles per hour and hence pulley would have linear velocity of 11 miles per hour

[somber coyote]

Harsh

[worldly niche]

and finally get revs per min from angular velocity

[short creek]

yea i got that i just was lazy

[minor dust]

Hello I need help with my trigonometry identities homework

Can Someone help me?

If someone can explain me that would be great

This is the homework

[rancid heart]

yall what does starts at and ends at mean when refering to bisectors and midsegments

[timber cargo]

tangent= sin/cos
csc=1/sin
cot=cos/sin
sec=1/cos

Use that to proof

[upper karma]

and also use Pythagorean identities

such as sin^2x + cos^2x = 1
sec^2x - tan^2x = 1
csc^2x - cot^2x = 1

use algebraic techniques to prove each identity

[sinful remnant]

Any reference angle must be an acute angle, right? So then this would mean that quadrantal angles do not have a reference angle?

[crimson pawn]

Yes

[smoky jetty]

i got the answer though it's not in the choices shared by the person. I used Pythagorean and the law of cosine. hmm

[upper karma]

So I just stared taking trig. My proffer said don't not try to remember the unti circle
, instead know how to find the reference angle. It will help you more down the road

Any truth to this??

[junior river]

The unit circle is great and more useful down the road imo

[silent plank]

learn the properties of the unit circle and how it relates to trig functions

do not remember every single value in that circle

[upper karma]

That makes sense

[silent plank]

knowing, reference angles (Q1) and properties of the circle/trig functions
allow you to derive the info you require as needed

[fathom cloud]

someone help pwease

[regal raptor]

if I’m right it’s 40+20

That equals 60

Wait

Hold on

Yea it would be 120

The missing angle equals 10

I believe

[gusty kernel]

a question from trigonometry ratio

[crimson pawn]

No

Wait

maybe

Ten wouldn’t make sense

Three angels to a triangle =180

Looking at the large triangle, the angles are 70 and 70

40° is both angles at the top

that anfle we are trying to find has to be greater than 20° but less than 40°

[regal raptor]

Yea I didn’t have connection I think I made a mistake

I didn’t account for 30

[crimson pawn]

Yea

[rancid walrus]

A plane is flying at a speed of 60 m/s and its vertical altitude increasing at a rate of 11 m/s. What is the angle of inclination of the plane to the nearest degree?

[valid comet]

hope this helps

[upper karma]

use opp/hypo so sin

sin x = 11/60

i believe so, right?

[waxen lotus]

Yes. Another way to think of it is y/r.
R is always the hypotenuse.
Y is the Y axis
X is the X axis

[rancid walrus]

I would've thought that the base would be the speed

Since you're flying horizontal not upwards

[atomic coral]

Looks 15

[upper karma]

yes

[valid comet]

If you were flying horizontal your vertical altitude wouldn't be increasing

[valid comet]

Do you know the side ratios of a 45° right triangle?

[young tusk]

do you guys know how to solve for x?

[fair oracle]

Is there a class past Trig? Like Trig II or whatever it is how it's Geometry -> Trig and Alg -> Pre-Calc -> Calc -> Calc II

[valid comet]

I don't think so

It's usually just trig functions in calculus n other subjects

[fair oracle]

Dang, alright

[upper karma]

This really has nice questions

[neon girder]

does anyone know the phase shit

[regal raptor]

the diameter of the circle is equal to the side lengths of the square

wait

never mind I’ve got nothing

[remote nimbus]

you can subtract the area of the circle of the area from the square.

[upper karma]

It’s ok I figured it out

[timber cargo]

You're right

[upper karma]

indian math olympiad!

[upper karma]

Yup.!

[potent fog]

0

There isn’t a phase shift in that eqn

[upper karma]

bro doing inmo lmao

hbcse official website

the same i saw there i guess

[upper karma]

So?
🙄

[upper karma]

bro, what are these questions? my head is dying?

Interesting tho.

[upper karma]

This is the beauty, lmao.!

The 6th one is actually awesome ngl

Tho after u guyz try it, u can check the soln..

[pliant roost]

yeaaaah, unlike the first ones it really is lovely

[upper karma]

Exactly

But all the questions are nice overall.!

[upper karma]

i need help with geometry.

like from the start

idk where to study for it tho

[timber cargo]

I suggest going to Khan Academy

[upper karma]

kk

i was thinking of that

[sinful remnant]

Does anyone know the why of the definition of trigonometric functions of any angle?

[remote totem]

how do i find the angle of theta when x,y and a,b is given (please give examples or something)

[wise pawn]

you can use the dot product formula

u dot v = |u||v|cos(theta)

[valid comet]

if you have the coordinates for the two points you could do tan^-1(y/x)-tan^-1(b/a)

[sour lynx]

I need help I dont know how the square root was answered

[upper karma]

can someone help me here?

[near hemlock]

The number of solutions for tan x + sec x = 2cos x in [0, 2π) is?

please helping

[zenith chasm]

multiply equation by cos x

then we can get 1+sin x=2cos^2 x

then (2 sin x-1)(sin x+1)=0

so there can be $\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}$

[zenith chasm]

use the double-angle formula: 2sin(arcsin(2w))cos(arcsin(2w))
then we can get
$4w \sqrt{1-4w^2}$

[somber coyote]

Nylch

Nylch

[lucid field]

I need to found out names for some shapes, I don't know how to describe them mathematically but I have screenshots of them

[dim mauve]

Does anyone know the trig property that makes the top equation equal to the bottom?

[wicked narwhal]

simpson laws

[dim mauve]

I searched that phrase on Google and WolframAlpha, but I didn't find anything. Do you have a source with more information?

[dim mauve]

Thank you.

[wicked narwhal]

in wikipedia, search "List of trigonometric identities"

[wicked narwhal]

I'm sorry I think it is better to search "sum to product formulas" or ''Werner's formulas" if you want more information

[upper karma]

I believe I found something wrong in my text book

[lucid field]

For example.
Does such shape have its own name?
it is symmetrical

[dark sparrow]

is it constructed as a cube with part of it sawed off

[timber cargo]

Its name is of no importance when I can describe it beautifully with dimensions

[hearty maple]

anyone who can help me with a tough conic section problem's soln, pls dm me

[wicked zephyr]

How am I just learning about this now ..

It’s really amazing stuff 🤩

https://tenor.com/view/sin-cos-einheitskreis-winkelfunktionen-math-gif-10679461

[smoky jetty]

thazz all neat!

[smoky jetty]

a really cool way to visualize the functions

[wicked zephyr]

Yes, 90 degree combined angles at origin, and 90 degree combined angles at circumference. Easier to see a trig function it’s cofunction always equaling out like this

[open swift]

hi

can anyone help me

@everyone

[timber cargo]

Did you seriously try to ping over 100 thousand people?

[weary drift]

@open swift everyone pings dont work here. but pls dont do it anyway

also just post your question

[open swift]

kk

[late echo]

Thanks to Pythagore we can solve this

And classic trigonometry

[upper karma]

Yo, are there any tricks for graphing sinusodal waves I am having a hard time with this https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/graphing-sinusoids/e/graphs-of-trigonometric-functions

[valid comet]

like horizontal shifts specifically?

[upper karma]

yeah

[valid comet]

are you having trouble identifying horizontal shifts, or graphing them yourself?

[upper karma]

IS THERE

geometry and trig

on sat math

[valid comet]

yes

usually triangles

[prime mortar]

Please help me on this

[topaz mountain]

how can I prove YZ is congruent to WV

i'm forgetting the reason

[silent plank]

how can I prove YZ is congruent to WV
you can't, there isn't enough info

[azure crown]

yes u can

[silent plank]

why do you think you can

[prime musk]

No u can’t…. U can prove similarity

[rugged frigate]

Angle VWZ= Angle WZY (WV parallel to YZ and co-interior angles)
Angle WVY= Angle ZYV(WV parallel to YZ and co-interior angles)
Angle WXV is equal to ZXY.
Hence by Angle Angle Angle Triangle VWX is similar to Triangle XZY
They are similar.

[nocturne remnant]

AAA proves similarity, not congruence

[prime musk]

That’s what I told u can only prove similarity from here

[rugged frigate]

Ah yes sorry my bad

[upper karma]

that is called being a true sigma. lol

[timber cargo]

I don't know what that means

[upper karma]

good afternoon everyone, I'm not sure how to show what this question is asking me. could anyone point me in the right direction?

[nocturne remnant]

just apply some basic geometry facts it seems like

e.g. the alternate angles are the same

so, to prove the desired bisection, you want to show that one of the triangles that form the parallelogram is isosceles

[dusk scaffold]

Idk how to do this properly

Need help

[dusk scaffold]

plz

[upper karma]

its just a joke regarding a ongoing interne trend.

[jolly sonnet]

Prove 🔺 EAC and 🔺DCB similar by AA
angle E = angle D ( each 90)
angle A = angle C (opposite angles are equal to opposite equal sides)

Now if they r similar
U can get the relation AE/DC = CE/BD

HENCE PROVED

[nocturne remnant]

Bro said hence proved at 200 db

[smoky jetty]

cool!

[dire snow]

Prove that, if in a triangle an exterior angle is congruent to the adjacent interior angle, then the other interior angles are acute. How can I prove this

Just the way

[smoky jetty]

not so sure but this is how I understand it:

Interior angle and Exterior angle of a triangle are supplementary (i.e they sum up to 180)

If the Interior angle is congruent with the Exterior angle, then each measures 90 degrees, thus the remaining 2 interior angles of the right triangle are acute?

Is that how? haven't done such thing yet

but i think the diagram is smth like this?

[dire snow]

Yeah

[upper karma]

hey guys

im scared to ask

but can someone explain sine cosine and tangent

[dire snow]

You Cna continue

[upper karma]

im in 8th grade (im 14)

and according to my math teacher

u learn trigonometry in 10th

grade

[dire snow]

Didn't do that sry

[upper karma]

and i always was ahead in math lol

and i can do everything

but i dont really understand trigonometry

[solemn ivy]

Prove that angel BCF=angel ADE

[graceful quest]

just prove adb simi to acb

similarity

[solemn ivy]

by what?

[graceful quest]

ab=ab

common

ad=bc given

as side are in equal ration

tri abd is similar to tri abc

ez

[dire snow]

. Does somoene know?

[solemn ivy]

in SSS?

[graceful quest]

yes

[solemn ivy]

but what is the third side

[smoky jetty]

another thing about an isosceles trapezoid's properties is its diagonals form a pair of congruent scalene triangles

[dense willow]

Guys how to find area of trapezium by bases and diagonals?

[smoky jetty]

whats the problem? can u share it?

[dense willow]

Oh yes i will write because it is not in english

Find area of trapezium, if bases are 6 cm and 9 cm and diagonals are 13cm and 14 cm

[grave pond]

Cut a length corresponding to the shorter base out from each of the bases. That makes the shorter bases disappear completely, and what you have left is a triangle you can solve to find the height of the trapezium.

Whoops, no, that would be if you have the sides rather than the diagonals.

[dense willow]

Ye

[grave pond]

If you have diagonals, then slide one of the diagonals to the right by the length of the shorter base. That makes its endpoint coincide with the endpoint of the other diagonal, and then you have a triangle with side lengths (diagonal, other diagonal, sum of bases) that you can solve.

[smoky jetty]

can u show an illustration?

[dense willow]

There is no illustration in the book

[smoky jetty]

i mean to tropo

[dense willow]

Oh okay

[smoky jetty]

im tryna see it, lol

[smoky jetty]

https://www.khanacademy.org/math/geometry-home/right-triangles-topic/intro-to-the-trig-ratios-geo/v/basic-trigonometry --the best I can do

[grave pond]

Suppose we want to find the trapezium ABDC. If we draw DE parallel to CB, then triangle ADE has side lenghts AD (known), DE (=CB, known) and AE (=AB+CD, both known). This lets us find the perpendicular distance between the bases, namely the altitude from D in ADE.

(You don't even need to use trigonometry: Heron's formula will give you the area of ADE, and then dividing by (AB+CD)/2 tells you its height.)

[smoky jetty]

yeah heron's