#Trigonomotry 2d triangles GR 11

87 messages · Page 1 of 1 (latest)

robust briar
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I can't figure out A) because I dont think i can split that triangle in half since it isnt proportinate. I have the answer sheet and the answer is 3.3 for its radious. Im waiting for my teacher to email me back but its been a while please help!

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wind marshBOT
late pike
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What rules are you familiar with for working with triangles

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There's a certain law that's helpful here

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As well as something we'll have to recognize from the scenario given

robust briar
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sine law wouldnt work, but cosine i require 2 sides length + angle

late pike
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Good

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We have an angle

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How many sides length do we have

late pike
robust briar
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1 side length

late pike
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Right, we know one

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But since the other two sides go from the center of the circle to the edge of the circle

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What must be their length?

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What's the distance from the center of a circle to its edge

robust briar
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2?

late pike
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Not necessarily 2

late pike
robust briar
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radius

late pike
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Yeah

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So we know the other two sides have length r

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So construct the law of cosines equation, and then solve for the radius r, since that's what we were looking for

robust briar
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oki i try it out

formal holly
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@robust briar

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You can split

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Do you know a bit of trigonometry?

robust briar
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rn im left off with 4squared = 2rsquared -2rsquared (cos 53) but doesnt the two r's cancel out? then theres nothing to solve

robust briar
formal holly
late pike
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Should be $r^2 = 4^2 + r^2 - 2(4)r\cos(53^\circ)$

robust briar
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riiight

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ill try again

formal holly
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@robust briar

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Bruv ig u made an error

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You won't get quadratic stuff in the first place

silver wadiBOT
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Coolempire93

470695562337976340.png
formal holly
late pike
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Yeah the quadratics cancel in here so you just end up with r = directly

formal holly
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Ye

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Like using the theorem of chord from centre bisecting another chord will be perpendicular and then just use trig

robust briar
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rsquared = 16 + rsquared -8r (cos53)

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:(

late pike
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Good

formal holly
late pike
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\begin{align*}
r^2 &= 4^2 + r^2 - 2(4)r\cos(53^\circ) \
&\text{subtract r^2 from both sides} \
0 &= 16 - 8r\cos(53^\circ) \
-16 &= -8r\cos(53^\circ) \
&\text{divide both sides by $-8\cos(53^\circ)$} \
\frac{2}{\cos(53^\circ)} &= r \
r &\approx 3.323
\end{align*}

formal holly
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Bruv how did you learn LaTex

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The lang so hard

formal holly
late pike
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Mm I forgot a newline

silver wadiBOT
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Coolempire93
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470695562337976340.png
short ravine
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you can just solve this using ASA because BC=CA since they are both radius of the same circle, so angle B = angle A

formal holly
late pike
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Angle B does = angle A but what does that do for radius

robust briar
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I got it !!!

short ravine
formal holly
robust briar
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ty ty

formal holly
robust briar
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I was never taught to use cosine that way

formal holly
short ravine
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the radius is $\frac{4\sin53}{\sin74}$

silver wadiBOT
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Helcovich Emire

1427447905005867008.png
late pike
short ravine
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using law of sines, 4/sin(74) = r/sin(53)

late pike
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Ahh law of sines

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Yeah

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That's a good one

formal holly
formal holly
robust briar
formal holly
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It's a rule

short ravine
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the central angle must be 74, because the other 2 angles are 53, so 53+53+74 = 180

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the central angle is the angle at the center of the circle

robust briar
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cuz its isococles?

short ravine
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so angle C

short ravine
formal holly
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Alr man ima roll out now , cyaa :)

robust briar
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oh, I seeeee yeah

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okay ty tguyss

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i might come back tho..... but lets hope not

late pike
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<@&268886789983436800> not sure if this is allowed but it feels wrong

zenith ridge
robust briar
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