Question

Consider a triangle ABC like the one below. Suppose that a = 35, b = 34, and c = 28. (The figure is not drawn to scale.) Solve the triangle.Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.If there is more than one solution, use the button labeled "or".1. B = 1: C = 0NOsolutionxХ5?CheckSanta

Answer 1

We are given a triangle with the following three sides.

a = 35

b = 34

c = 28

We are asked to find all three angles of the triangle.

The angles of the triangle can be found using the law of cosines.

1. Angle A

$\begin{gathered} \cos(A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos(A)=\frac{34^2+28^2-35^2}{2(34)(28)} \\ \cos(A)=0.3755 \\ A=\cos^{-1}(0.3755) \\ A=67.9\degree \end{gathered}$

So, angle A is 67.9°

2. Angle B

$\begin{gathered} \cos(B)=\frac{a^2+c^2-b^2}{2ac} \\ \cos(B)=\frac{35^2+28^2-34^2}{2(35)(28)} \\ \cos(B)=0.4352 \\ B=\cos^{-1}(0.4352) \\ B=64.2\degree \end{gathered}$

So, angle B is 64.2°

3. Angle C

$\begin{gathered} \cos(C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos(C)=\frac{35^2+34^2-28^2}{2(35)(34)} \\ \cos(C)=0.6710 \\ C=\cos^{-1}(0.6710) \\ C=47.9\degree \end{gathered}$

So, angle C is 47.9°

Summary:

A = 67.9°

B = 64.2°

C = 47.9°