what
#geometry-and-trigonometry
acoustic jungle
gray marten
I have no idea how to do this
glad gorge
Is dis part 60 ?
gray marten
oh yea probably i think
humble pebble
yea
glad gorge
Oh :0
humble pebble
and other is 90
silent plank
the highlighted black part? no
glad gorge
Ohhh
humble pebble
point of tangency
gray marten
Oh yeah lol
glad gorge
Tysm
gray marten
60+90+x=180
humble pebble
just add and subtract from 180
gray marten
soo y = 180-60-90 = 30
silent plank
wtf is x?
humble pebble
30
gray marten
ye soz lol i meant y
glad gorge
...
glad gorge
Isn’t a triangles sum suppose to be 180
silent plank
that is true. but you are improperly applying other theorems
acoustic jungle
I was talking about the circle.
acoustic jungle
I'm not sure how to do it without inscribed angle theorem, but I would use inscribed angle theorem to find y.
glad gorge
Oh
gray marten
okay let me find out what that is lol
silent plank
it may be easier to visualise if you connect the center to that vertex
glad gorge
Uh
silent plank
*makes it easier to see how to apply the inscribed angle theorem,
as well as an alternate method
indigo field
someone help
glad gorge
Alright thx ^^
gray marten
So was y = 60?
upper karma
@indigo field do you know how to get the sum of all the measures in any polygon?
indigo field
no 😹
upper karma
180(x - 2)
X = amount of sides
So plug it in
5 sides
So 540
Now add all the angles together set equal to 540
And find x
Using simple algebra
And then plug it in for all of them and find the smallest angle
@indigo field u got it?
indigo field
no im lost
upper karma
180(x-2) is the equation for finding the sum of the angles in any polygon
X is the amount of sides
indigo field
yes yes i’ve got that
upper karma
You just set them equal to 540
And use algebra
To find x
https://youtu.be/7Cx9s5cKnDk this is a pretty good video
Seems like your question
@gray marten yes?
gray marten
uh so, the other persons question, is it safe to say that putting that line there makes an equilateral so y = 60?
high zephyr
not sure, looks like isosceles to me
gray marten
rip
upper karma
@high zephyr how so
high zephyr
the radius
high zephyr
i said im not sure
high zephyr
xi, mind explaining why u said it's equilateral?
silent plank
equilateral triangles are also isosceles
high zephyr
yea the 2 radii but what about the third side
upper karma
Idk
silent plank
top angle is 60°
upper karma
Oh ya
60 60 60
It’s equilateral
@gray marten
Central angle = arc that intercepts I
It*
You know the central angle = 60
Based on the 2 radii you can say it’s isoceles
But since the top angle is 60
The other 2 angles must be 60 too
high zephyr
the angle at centre = 60 then
gray marten
Noice, so yeah y = 60
Wanted to ask, how one would go about finding the area of this geometrically.
It seems that you can find the area of this really nicely using linear algebra.
by finding the determinant.
But I wondering if anyone had an idea of using more geometric ideas to get the area.
Area's 5 btw.
storm lily
I mean, I suppose you can consider the triangle formed by (0,0), (3,0) and (3,4) which is ½ × 3 × 4 = 6, then subtract the triangle formed by (0,0), (2,0) and (2, 1) = ½ × 2 × 1 = 1, then subtract the square formed by (2,0), (3,0), (3,1) and (2, 1) = 1 × 1 = 1 and the triangle formed by (2,1), (3,1) and (4,1) = ½ × 1 × 3 = 1.5, so this is half the figure, so the total area is 2 × (6 - 1 - 1 - 1.5) = 2 × 2.5 = 5?
gray marten
Oh nice that works too, yea
upper karma
Use the box method
And subtract all the triangles
From the area of the square
Made
novel flax
ya
or you could use coordinate geometry and distance formula and drop a perpendicular to find a height then calculate area that way, as an alternative solution
gray marten
Yea that waht I was thinking tooo
Or even easier, I think you can just use pythagoras
for height too.
Use pythagoras to get b, sqrt(2^2+1^2) = sqrt(5)
And for height, sqrt(1^2+2^2) = sqrt(5)
So A = sqrt(5)*sqrt(5) = 5
novel flax
the parallelogram looks like 2 right triangles combined, if you draw the diagonal
how did you get that @gray marten
what is b and h
gray marten
Area of parallelogram is A = b*h
Although I assume, h isn't always between the two points, it seems like it is for this one.
novel flax
yes
gray marten
Though as xi, suggested, box method seems pretty good too
Since you'd just need to do 4*3 - (3/2*1)*2 - (4/2*1)*2
marble parcel
Can someone help me with these?
somber coyoteBOT
The following error occured while calculating:
Error: Syntax error in part "\frac{4}{3}(\pi)(\frac{6.32^{3}}{8}+(\pi)(\frac{6.32^{2}}{4}(17.5)" (char 1)
vital bough
@acoustic jungle thank you
high zephyr
@vital bough my fault for misleading u
vital bough
it is okay happens too all of us
foggy beacon
hey guys I have a question here if someone can please help me clarify it will be greatly appreciated
for the cos i found that the angle was 114.6 degrees, for the ratio for sin, i found that the angle was 65.27 degrees. does that mean question C can have two solutions
and for tangent i got -65.35 degrees
but like i don't think the 114.6 degrees angle is related to the 65.27 degrees
should i just give off all of those answers or should i just give the cosine angle alone
upper karma
There are two solutions
foggy beacon
oh
foggy beacon
what about the tangent ratio tho?
because in question b i had to get all of the ratios for the primary trigonometric ratios
so should i give three
upper karma
thats weird
Cos=-5/12
literally
should
have 1 answer
wwait
im guessing you can have a positive angle
and a negative angle
foggy beacon
-245.4?
foggy beacon
no for tangent i got -65
foggy beacon
should i just answer for all the angles for tan, sin, and cos from those ratios ?
i got 114.6 degrees for cosine
should i do 360 - 114.6?
upper karma
Just a heads up, cos(x) = cos(x+2k*pi) where k is integer
So Im guessing there could be more than 2 solutions
shouldnt
cos144.6
be
-5/12
so the angle is fine
do 114.6-360 if u wanna find the same angle that is coterminal
foggy beacon
okay thank u bro love u guys thank u for the help
glass crag
Hey I was just doing some work and I got through all of it but got stumped on this question. My usual study group isn't around to help and yea, this place is my last resort. Anyone able to help? (I've never worked with a 90, 70, 20 triangle before)
(also first time working in third dimensions)
upper karma
@glass crag use trig function to calculate height and then just calculate the area length times width times height
Use trig function to find other leg of triangle
And that’s basically the height
And then calculate area
glass crag
Alright thank you!
upper karma
Your welcome
foggy beacon
hey guys i have a very quick question
right here like i proved this identity but i just dont know if that's what the question is asking
do i prove it first then plug the 30 degrees in this identity
dark sparrow
yeah it looks like a badly worded way to say "show that $1 + \cot^2(30\dg) = \csc^2(30\dg)$"
somber coyoteBOT
Ann:
gray marten
what happened to ur lovely homestuck pic Ann
versed river
thats impossible
upper karma
^
versed river
unless theres a right angle, or something like that
upper karma
@lost vector post us the real problem
lost vector
Emm
dark sparrow
are you given anything else besides the lengths 10cm and 7cm
if not then the problem is impossible
lost vector
Ok wait
dark sparrow
the most you could say is that AC has to be between 3 cm and 17 cm to satisfy the triangle inequality
but beyond that, nothing
lost vector
Wait
dark sparrow
oh NOW we're talking.
dark sparrow
what's the vertex on the very left there?
lost vector
It ok
lost vector
Ye
dark sparrow
ok
well
and that circle is tangent to all the sides of the triangle
so then you have KA = KB, LA = LC and MB = MC
lost vector
Thanks
upper karma
@lost vector that was a whole different problem
lost vector
Ok ok
upper karma
Np
cobalt bear
sly marlin
@cobalt bear when the ship is closest the pier is directly to the left or right of the ship
novel mango
i have a triangle with arbitrary sides a,b,c with a>=b>=c and with area 1 i have to prove that b^2>=2
so if i draw a perpendicular p to the side b area =2=b.p how do i prove that b^2>=b.p
or b>=p
<@&286206848099549185>
twilit zenith
Hint, to God V2:
||Draw the triangle, and try to check what happens if the reverse is true. Meaning, if we have b < p, what can you say about either a or c?||
hollow hazel
@KanaoTsuyuri just find the perpendicular distance of water surface and point P, that will be the shortest distance using sin(30°). In second problem ship will travel 40km so it will not be close
acoustic jungle
?
upper karma
@upper karma I tried to solve it last night using the Law of Sines/Cosines by assigning a length of 1 to one of the edges
here's what I had, but keep in mind that I was half-watching a movie at the same time:
x = 20 degrees
angle CDB is also an isosceles triangle (20 deg on one side, 20 deg on the other), so edges CD and DB should be equal
except mine were off by 0.001 😛
but they were close enough so I rounded off the angles
what's weird is that the verification process gave me angle EDP as having 70 degrees (instead of 110 what I wrote), not once, not twice (both crossed out in the picture above), but 3 times
and that angle also looks (visually) like it might be 70 degrees
110 would make it look wider than 90, which in the picture isn't
so I was most likely wrong 🙂
upper karma
click "open original"
or remove the "width" and "height" part from the link: ...50032/unknown.png**?width=267&height=399**
acoustic jungle
do you want to solve it the trig way or the online way.
upper karma
what's the online way?
acoustic jungle
nevermind
it's a bit different
I thought they were the same thing oops.
It might help though, they are really similar
acoustic jungle
@upper karma lol for the 70 degree thing it's because sin(180-x)=sinx
So when you are arcsining, you are getting 70 instead of 110
upper karma
if it looks like a 70 degree angle, and quacks like a 70 degree angle... it's probably a 70 degree angle
upper karma
110 adds up with the other angles
upper karma
I'm gonna declare x as 20 degrees and call it a day
are you getting this, @upper karma ?
once again, some random Indian guy on Youtube saves the day
and I was right 😛
20 degrees
but I'm gonna go toot my own horn somewhere else now 😄
acoustic jungle
Yeah. The reason you verified that you got 70 instead of 110 is because sin(x)=sin(180-x)
I was saying 110 is right and 70 was not.
upper karma
@acoustic jungle and how do you know when to use one over the other?
acoustic jungle
Well you don't, but the way you got x=20 should be correct, so that meant it's 110.
I was just saying when you were verifying, the reason you got 70 instead of 110 is because of the reason I said above.
upper karma
I understand
I had to look it up why, tho
and this post helped: https://www.quora.com/Why-is-sin-θ-sin-180-θ
from what I remember from Khan Academy lessons, if a triangle has an obtuse angle, you use 180-θ
because you can't always rely on visual cues
for example:
one is clearly obtuse
sometimes they stretch or squish those figures like that intentionally, to throw you off
always rely on the numbers
numbers don't lie
humble pebble
this is a pre weird question but i cant draw something that fits
mainly this where i can measure them all; The logo must contain at least one central angle, at least two inscribed angles and two chords
upper karma
which point(s) are you having trouble with?
humble pebble
simply drawing a logo where it has a one central angle, at least two chords, and two inscribed
upper karma
do you have a protractor?
upper karma
it can be parallel to one of the radii
that would be the relationship between them
they're both parallel to each other
upper karma
or they could share a common point on the circle
humble pebble
ah
upper karma
if they share a common vertex, sure
pulled literally from the first 4-5 images by searching "inscribed angle" on google
upper karma
description says at least one
you can use more than one
@upper karma I tried to solve it last night using the Law of Sines/Cosines by assigning a length of 1 to one of the edges
@upper karma cool, I was moving toward the same xD, but the video that @acoustic jungle shared here was much easier.
humble pebble
yes @upper karma but then I need to put another inscribed angle, and then when i add a chord, because the way it's drawn it's impossible to find the chord
upper karma
we had this problem during high school i guess, but it's been totally forgotten
silent plank
wdym impossible to find the chord?
upper karma
silent plank
the question doesn't say everything needs to be connected
i.e you can literally connect two points on the circumference and you'll have +1 chord
humble pebble
how will i measure that
silent plank
ruler,protractor, and/or formulas
upper karma
do any of them form a right angle with that diameter?
did you measure any angles at all?
(point #4)
you can also use a glass or cup to draw a circle instead of eyeballing it
or a drawing compass
be careful if you use one of those, to not punch a hole in your copybook
use a plastic ruler or something behind the page
or maybe just don't press that hard
viscid ginkgo
how can I solve sin a - sin b = 2cos((a+b)/2)sin((a-b)/2)
a = 30 degrees, b = 90 degrees.
zenith wave
can anyone help me_
silent plank
where did you get stuck?
zenith wave
Actually i figured that one out can i gat help in this?
i dont know how to get the 2x-y2 to the other side
silent plank
consider getting all your variables to one side and complete the square
zenith wave
this is to find the center and radius of the circle btw
upper karma
i dont know how to get the 2x-y2 to the other side
@zenith wave to get that term to the left itd go as
$-(2x-y²)$
somber coyoteBOT
Al3dium:
somber coyoteBOT
Al3dium:
upper karma
@zenith wave
zenith wave
i got you
high zephyr
do u need to solve for a or b?
upper karma
I think both of them.
humble pebble
a and b are not inside the triangle
humble pebble
idk maybe degree comes in but it's 2 squared plus b squared equals a squared
uncut mist
Any reason 2 is not called c ?
humble pebble
2 is not c
c is the hypotenuese
the longest side
90 degree points to the longest side
a and b don't matter they're interchangeable
upper karma
yea, I need progress to finish but idk how to solve it
I was expecting like some kind of formula
high zephyr
do u know ur trig ratios?
upper karma
nop
high zephyr
sin, cos, and tan
ok
sinx=o/h, cosx=a/h, tanx=o/a
where o=opposite, a=adjacent, and h=hypotenuse
opposite is the side opposite the angle
hypotenuse is the longest side of the triangle so in this case it's the slanted one
adjacent is the side next to the angle
an easier way to memorise this is SOHCAHTOA
upper karma
ah okay, I searched it up and god a clear image.
high zephyr
for the ratios, u just sub in the known angles and sides to find the unknown ones
obtuse hornet
no one answered me tho
obtuse hornet
whats the asnwer
upper karma
Search up the law of cosines equations and then just plug it in
Cos(theta) =
...
U look it up
high zephyr
@obtuse hornet
upper karma
Ya
high zephyr
sub the known sides in to find the angle
upper karma
Yep 👍
high zephyr
for convenience https://cdn.discordapp.com/attachments/326138757474680852/702284616895299724/unknown.png
upper karma
A^2 is always the opposite side of the angle ur trying to find
Cos(theta) = 21^2 + 14^2 - 13^2/2(21)(14)
You could also use it when u have sas
obtuse hornet
Ok I understand, but i dont got paper or anything so whats the answer haha
upper karma
And ur trying to find the last side
Ok
Wait
@high zephyr u do it
I’m out
Cos(theta) = 21^2 + 14^2 - 13^2/2(21)(14)
It’s that
high zephyr
^ answer
upper karma
@obtuse hornet just plug that in to ur calculator
high zephyr
we're helping u get the answer, not giving u it
but since xi already said it...
upper karma
After you find it you use arc cosine btw
obtuse hornet
where did the theta go
obtuse hornet
the cos is theta?
upper karma
Just use inverse cosine
Forget what I said earlier
Theta = arccos(21^2 + 14^2 - 13^2/2(21)(14))
acoustic jungle
upper karma
,ask Theta = arccos(21^2 + 14^2 - 13^2/2(21)(14))
somber coyoteBOT
Requested by Al3dium#4084
acoustic jungle
upper karma
Where tho
acoustic jungle
upper karma
Im not sure wait
Theta = arccos((21^2 + 14^2 - 13^2)/2(21)(14))
Now
,ask Theta = arccos((21^2 + 14^2 - 13^2)/2(21)(14))
upper karma
@obtuse hornet
acoustic jungle
upper karma
Still?
high zephyr
answer should be around 37 degrees
obtuse hornet
what does that mean
upper karma
Lmao i took it out of context
acoustic jungle
that is obviously wrong
acoustic jungle
because he didn't use
upper karma
Where?!?!
What
(insisting on being out of context)
Did I do something wrong
Sorry for my mistake
Np
upper karma
,ask arccos (39/49)
obtuse hornet
;-;
upper karma
Uhh
acoustic jungle
What are you guys doing
upper karma
Since i joined, i only copied and pasted the messages i see to ,ask lmao
high zephyr
ignore the rest
upper karma
Yeah
cobalt scroll
can someone pls help me with this question... 4cot^2(θ)-8=2cosec^2(θ)-5cosec(θ)
<@&286206848099549185>
high zephyr
^ already asked
high zephyr
area of sector - area of triangle
acoustic jungle
it depends on your angle.
humble pebble
yeah, I just subtracted the triange, and subtracted 2 halfs of the central angle
zenith wave
can anyone help me?
viscid ginkgo
hello I am having trouble using the product sum formula
acoustic jungle
AC^2=OA*AB
acoustic jungle
I just answered your problem.
zenith wave
Can u explain how you did it
acoustic jungle
it's called the secant tangent formula.
zenith wave
How do i find the length of segment AB tho?
upper karma
Since you know a tangent to a circle has a 90* angle
You can classify that as a right angle
Then just use Pythagorean theorem
Radius is 3 units
3^2 + 4.5^2 = AO^2
What ever AO is u subtract by the radius 3 units
And U get AB
@zenith wave That’s another way to do it
zenith wave
got you
upper karma
Your welcome
acoustic jungle
whoops lmao
hello I am having trouble using the product sum formula
I thought that was you and I thought that meant you wanted to use the tangent secant formula
yeah the right triangle way is much better.
upper karma
👍
prime jewel
Hello guys, how do i know if a triangle has 1, 2 or 0 solutions.
In that example they get the angle A
any ideas?
devout harbor
wdym by 1,2 or 0 solutions? In a triangle there are 3 angles, 2 are given to you, you found the third.
prime jewel
It's a weird problem
Yo need to find 2 possible solutions
because the number you get SinA= number
Can be 2 different angles
devout harbor
we can only find one solution because sum of angles = 180
glacial haven
hes asking if its an ambiguous case
dark sparrow
what's giving you trouble here?
dark sparrow
you don't need any identities here
all you need is to solve the equation $\sin(\theta/2) = \frac{6}{11}$
somber coyoteBOT
Ann:
upper karma
i'm not really sure how to go about doing that
dark sparrow
ever heard of arcsin?
versed river
,w sin(2x)=6/11
somber coyoteBOT
Requested by Sneaky#0928
versed river
ok yes its correct
dark sparrow
okay where are you stuck now
upper karma
you have to use the unit circle for this one correct?
dark sparrow
it can be done 100% algebraically but a unit circle might help you more easily pin down what quadrant alpha/2 will be in
so that you can get the signs right
upper karma
okay, so i can pinpoint where pi/2 is and how cos is X on the unit cicle, not really sure how to figure out the 4/5 part
dark sparrow
overthinking.
you're overthinking it
from the unit circle, all you really need is that α/2 will be in the first quadrant and hence that its cosine will be positive
then you can use the identity $\cos^2(\alpha/2) = \frac{1 + \cos(\alpha)}{2}$
somber coyoteBOT
Ann:
digital gulch
Ive been stuck forever
quiet mason
volume_{entire cone}=2*volume_{cone with centre of base B}
directly follows from the fact that yoou dvide it into two parts wIth equal volunes
digital gulch
Yeah that makes sense
But how do i approach finding the ratio of the area of circle B to the area of circle C
quiet mason
you can find ratio of the radiuses
quiet mason
areas are just a ratio of (r1/r2)^2
you can try doing some questions
if you get stuck you can post
vague berry
got it
quiet mason
nice
digital gulch
I dont know how to approach it
dark sparrow
let V, r and h be the old volume, radius and height of the cone and let V_1, r_1, h_1 be the new volume, radius and height of the cone
$V = \frac13 \pi r^2 h \ \ V_1 = \frac13 \pi r_1^2 h_1^2 \ \ h_1 = 0.92h \ V_1 = 0.85 V \ r_1 = (1+\tfrac{k}{100})r \ k = ; ?$
somber coyoteBOT
Ann:
dark sparrow
@digital gulch if you're still here
acoustic jungle
oh oops.
eager wasp
What are these questions asking me to do?
dark sparrow
no, fishraider
acoustic jungle
the largest value 3sinx can be is 3 and that's when x=90
dark sparrow
yeah but the way you said that earlier is confusing as hell
acoustic jungle
oh sorry. 🍉
eager wasp
Whaaa
dark sparrow
(sorry, i have to go again.)
granite elk
Anyone knows how to solve this?
i know one of the answers is 3.3 from my calc but im not sure how to find the others
how do i find the values?
silent plank
have you learned about general solutions yet?
silent plank
i'm assuming that since you got 3.3 that you had:
$$\tan(3\theta+20\deg) = \frac{1}{\sqrt{3}}$$
at some point
somber coyoteBOT
ramonov:
granite elk
ye
silent plank
here, you should consider the general solution for tan
$3\theta +20\deg = 30\deg + k\cdot 180\deg$ where $k \in \mbb{Z}$
somber coyoteBOT
ramonov:
granite elk
silent plank
yeh. that.
granite elk
thanks
But what do i do following that to get the other solutions
it has been a really long time since i did that type of work so im pretty lost
granite elk
like 30,60...?
granite elk
so everything from 0->180
silent plank
$3\theta +20\deg = 30\deg + k\cdot 180\deg \
\theta = \ ?$
somber coyoteBOT
ramonov:
granite elk
theta=(10+k*180)/3
silent plank
it's better if you split the fraction here
silent plank
$\theta = \frac{10\deg}{3} + k\cdot 60\deg$
somber coyoteBOT
ramonov:
silent plank
and then plug in appropriate values.
i.e. 0,1 ... etc. (since k is an integer)
until the value you get is outside the interval
granite elk
ohhhh
ok thank you so much
so my answers would be what ever 0 1 and 2 give me
im pretty clueless about this its been so long
thanks
silent plank
tan is quite nice to work with.
the general solutions for something like sin are little more complicated so practice more
granite elk
Just to make sure
if i was given
in the 0<=x<=180
and i wrote
which makes x=75
what do i do with the 75 to find an answer?
silent plank
same as before, consider general solutions
x=75° will only be one of your solutions
granite elk
i think im supposed to do 360-75
granite elk
i think 75 might be the only answer?
rich wolf
He said 0<x<180
silent plank
how would you apply that to this question?
granite elk
silent plank
2x = ?
granite elk
premise of the question
is that a student tried to answer it
but made a mistake
his mistake is 360-75=285
but i dont even know why he subtracted 75 from 360 in the first place
silent plank
yes. the mistake was apply the properties of the cos function after already solving for x
granite elk
ah
silent plank
295° isn't even a solution to the original equation
silent plank
which is why you should be applying the general solution before isolating x
silent plank
yes, 285° would be one. but the work doesn't properly justify it
upper karma
Isnt it safer to apply double angles here
silent plank
$2x = k\cdot 360° \pm \arccos(-\frac{\sqrt{3}}{2})$
somber coyoteBOT
ramonov:
granite elk
have not covered arccos yet
silent plank
arccos is a much better way of writing cos^(-1)
silent plank
no. and in use
lime storm
sorry?
weary drift
channel's occupied
silent plank
that sounds right
@lime storm this channel just opened up. show your work and/or what you applie.
upper karma
Express these three vectors as a linear combination of the vectors i and j
and in algebraic form*
Would be the translation
I dont understand how to do this
pastel anchor
pastel anchor
yessir
wait but i need to find the arc
of AB
well nvm i found AB its 80
but idk about thee rest
upper karma
Try with inscribed angles theorem @pastel anchor
novel flax
ACB is an inscribed angle that intercepts arc AB because C is a point on the circle
pastel anchor
ok is DBC also inscribed angle?
novel flax
yes it is
upper karma
is there an inscribed angle in there?
@pastel anchor yes
pastel anchor
so the angle of inscribed angle is half of the arc
pastel anchor
so angle DBC is 45
upper karma
Wowowo
pastel anchor
because arc DC is 90?
novel flax
yes
pastel anchor
what do i do next
upper karma
Have you done b)
pastel anchor
its 40 right
upper karma
Yep
pastel anchor
how can i find angle P
upper karma
@novel flax you take over me?
pastel anchor
no
novel flax
the measure of angle P is one half the difference between the meausres of arc AB and arc BC
since those are the two arcs intercepted by angle P
pastel anchor
how can i find arc BC
upper karma
Okay the other guy got answered so i can kinda go back
Sl
@pastel anchor you should honestly look for theorems
This problem is based on theorems basically
Wait
pastel anchor
what is the theorem to find arc bc idk
upper karma
Why you want arc AB
pastel anchor
the measure of angle P is one half the difference between the meausres of arc AB and arc BC
upper karma
True
upper karma
Im thinking on another way of approaching this wait a sec
but i need arc bc
@pastel anchor you mean ac
Well yeah
Wait
pastel anchor
im confused
pastel anchor
i ahve to do it with secant tangent chords theorems and shit
novel flax
oh
upper karma
novel flax
i have a way to find the arc bc that uses tangent-chord angles but it involves trigonometry
pastel anchor
but why would my geometry teacher send work involving trig
novel flax
yes law of sines
novel flax
did your geometry teacher teach you trig
it seems like the most practical method since you're given 2 side lengths of a triangle
upper karma
Yeah
pastel anchor
he did cosines sines and tangent
upper karma
So then id think itd be valid
pastel anchor
but there has to be a way using the angle theorems no?
upper karma
Although you werent taught law of sines, right?
but there has to be a way using the angle theorems no?
@pastel anchor idk, but Orange and i only know to do it with law of sines
pastel anchor
yea i prob forgot, it was like sinle/tangent or w.e
upper karma
Oh then it might be it
upper karma
So yeah, you should try with law of sines
@novel flax could you post your approach with law of sines bc i did it on a whiteboard
Might be better exposed and more clear than on whiteboard
novel flax
sure
upper karma
@pastel anchor this one is the law we are talking about
He'll now post the application
novel flax
to solve for an angle you need to use inverse sine function, or arcsin
pastel anchor
no
novel flax
upper karma
Which is the same
novel flax
here is the solution above with trig
pastel anchor
alright thnaks
upper karma
Np
pastel anchor
novel flax
i had that same question for my geometry hw today
upper karma
No way lmao
novel flax
arc of AD is the measure of arc AC = arc CD
pastel anchor
but idk the measuer of arc AC
novel flax
arc AC and arc BC have the same measure because they are intercepted by congruent chords @pastel anchor
jovial tendon
can anyone help me with 5 simple geometry questions, wasnt in for the topic and for obvious reasons cant contact my teacher for help as schools are shut, if anyone wants to help can you message me, thanks
pastel anchor
@novel flax do you know what theorem is that?
novel flax
In a given circle, if two chords are congruent, then the corresponding intercepted arcs are congruent.
dense sky
upper karma
@dense sky quickly
Those are basically all congruent angles
So the sum of all of those angles should sum up to 360
dense sky
so something for x that will make em equal to 360?
sorry my teacher isn’t the greatest with google classroom so i’m pretty clueless 😅
dense sky
sure or you could always dm me
novel flax
subtract arc CD from arc AC
pastel anchor
idk arc CD though
novel flax
hint: angle CAD is an inscribed angle that intercepts arc CD
pastel anchor
BABHHH
ah
holy shit how do you notice that my brain is fried
wait but there is no angle for BDA, so how can i find arc BDA
novel flax
use tangent-chord angles
wait
arc BDA is the sum of arc BC and arc CA @pastel anchor
and you know those measures
pastel anchor
@novel flax so its 370? 135+135
novel flax
270, but why do you need it for
novel flax
cos 180+40 = cos 180-40 = -cos40 and cos^2 40 + sin^2 40 = 1
vague berry
wait is that for me?
novel flax
ya
vague berry
wait are u sure
novel flax
yes
upper karma
Theorems
vague berry
ok
It is known that sin(40˚) = A. Find the value of cos(220˚) in terms of A
so that would be just 1???
novel flax
no
novel flax
manipulate the pythagorean identiy to get
pastel anchor
what etheorem do i use to find angle BEC @novel flax
vague berry
oooo ok
novel flax
2 instead of @
novel flax
cos 220 = negative sqrt (1-A^2) @vague berry
intersecting chords theorem since you know arc BC and AD @pastel anchor
upper karma
Did I draw this right?
novel flax
@upper karma no that is not angle of depression
the other angle is supposed to be 20 degrees
upper karma
What have you got so far
pastel anchor
is BAF inscribed angle?
novel flax
that was the last question on my hw too
upper karma
Yeah
novel flax
but bf is parallel to ad
so arc ab and fd are congruent
so will be the intercepted inscribed angles
pastel anchor
how do we know they are parallel though wtf
novel flax
given
pastel anchor
i just saw
acoustic jungle
upper karma
Lmao
pastel anchor
i dont think i learned intercepted inscribed angles
novel flax
I meant the inscribed angles that intercepted those arcs
remember it's half the measure of the arc
pastel anchor
i still dont get i only have one angle so far and thats 20
pastel anchor
isnt arc BF 40? @novel flax
novel flax
yes sorry
novel flax
arc ab + bf + fd =180
pastel anchor
ive never heard of that theorem before. what is the theorem?
pastel anchor
oh wait semi circle?
novel flax
afd is a semicircle
pastel anchor
so arc AB is 70 and arc FD is 70?
novel flax
yes
pastel anchor
can we even use intersecting chords theorem to find angle BEF @novel flax
acoustic jungle
yes.
pastel anchor
don't we need the arc to on the other side
acoustic jungle
The other side is 180
pastel anchor
40+180/2 is teh equation?
acoustic jungle
Yes.
pastel anchor
do i use chord tangent theorem to find angle CDB
acoustic jungle
You can.
pastel anchor
what would u use
pastel anchor
how do u know adc is 90?
wait but that means angle CDB is 45, so when i do 1/2(110) i get 55
upper karma
@novel flax if you're gonna take a screenshot on your phone, don't make it a vertical screenshot, eh?
novel flax
ok
pastel anchor
send help mr orange @novel flax
acoustic jungle
that is 120/360 of the area of the circle.
acoustic jungle
correct
copper valve
Hm?
high zephyr
what
copper valve
wdym
high zephyr
all he did was help
copper valve
lets not use that word
upper karma
What ever circumference is
worthy igloo
what's your problem
acoustic jungle
are you good.
upper karma
Multiply by
120/360
And there
I think
Idk
Or it’s
The Formula is 1/2 R^2 x Theda
Find area
Idk
high zephyr
it's 120/360 times by area of circle
upper karma
I’m done