Question

Solve the triangle.

A = 33°, a = 20, b = 13

Answer 1

Are you assuming a right triangle?

Answer 2

SOH CAH TOA is coming I bet. :)

Answer 3

@eSpex not a right triangle

Answer 4

Note that: \[\frac{a}{\sin(a)}=\frac{b}{\sin(b)}=\frac{c}{\sin(c)}\]

Answer 5

Ah, an SAS solution, can't wait.

Answer 6

Use the Law of Sines, which states: Applying this law to the given situation, we have a triangle like this: So it follows that:\[\frac{ \sin(33°) }{ 20 }=\frac{ \sin(B) }{ 13 } \implies \sin(B)=\frac{ 13 }{ 20 }\sin (33°) \implies B = \arcsin \left( \frac{ 13 }{ 20 }\sin (33°) \right) \approx 20.733\] So we know angle A = 33° and angle B = 20.733°. Since all angles add up to 180°, angle C = 180 - ( 33 + 20.733) = 126.267°. Now we have all angles A, B, C and we have sides a and b. We just need to find side c which we can find once again by applying the law of sines. So now let's compare the ratio of sin(A)/a = sin(C)/c and solve for c since we know the values of angle A, side a, and angle C.\[\frac{ \sin(33°) }{ 20 }=\frac{ \sin(126.267°) }{ c } \implies c = \frac{ 20 \sin(126.267°) }{ \sin(33°) } \approx 29.608\] Therefore, the triangle ABC, has Angle A = 33°, Angle B = 20.733°, Angle C = 126.267° and has Side a = 20, Side b = 13, Side c = 29.608.

Did you understand that? You should also know that you can use the law of cosines here as well if you find that easier.

@Ambition

Answer 7

You appear to have your law of sines inverted.

Answer 8

Taking the reciprocal of all sides yields the same equality, mate.

Answer 9

@genius12 yes I do understand. How would you use the law of cosine

Answer 10

@eSpeX

This is how the Law of Sines works in case you didn't know:\[\frac{ \sin(A) }{ a }=\frac{ \sin(B) }{ b }=\frac{ \sin(C) }{ c }\] OR\[\frac{ a }{ \sin(A) }=\frac{ b }{ \sin(A) }=\frac{ c }{ \sin(C) }\] Doesn't matter if it's inverted or not, as long as the ratios you are comparing are consistent in terms of side/angle, the ratio retains.

Answer 11

but on the other hand, if you are solving for a length of a side, why put it in the denominator?

Answer 12

@satellite73 They know how it works.. :)

Answer 13

Good job all.

Answer 14

For law of cosines, you just plug in angle A, side a, side b, and solve for side c. This might be potentially shorter as a method of solving so it might help you. This is how it will go:\[a^2 = b^2 +c^2-2(b)(c)\cos(A) \rightarrow 20^2=13^2+c^2-2(13)(c)\cos(33)\]Rearranging will allow you to solve for c and then you can find the remaining angles using SOHCAHTOA or you could apply the law of sines or the law of cosines once again. There is many ways you could do it.

@Ambition

Answer 15

Thank you all for your help

Answer 16

No problem.