Chapter 13.5, Problem 11PPS

To determine

To find: To find the angles of the triangle to the nearest degree.

Answer to Problem 11PPS

In triangle ABC

Angle $\angle A={31}^{\circ }$

Angle $\angle B={108}^{\circ }$

Angle $\angle C={41}^{\circ }$

Explanation of Solution

Given information:

The length three sides of triangle ABC.

AB= c = 9

BC= a = 7

CA= b = 13

Calculation:

To find the angle A, B and C we use the cosine formula for any triangle:

  $\begin{array}{l}{b}^2=a^2+c^2-2ac\cos B\\ \Rightarrow {13}^2=7^2+9^2-2\times 7\times 9\times \cos B\\ \Rightarrow 169=49+81-126\mathrm{Cos}B\\ \Rightarrow 169=130-126\mathrm{Cos}B\\ \Rightarrow \mathrm{Cos}B=\frac{130-169}{126}\\ \Rightarrow \mathrm{Cos}B=\frac{-39}{126}\\ \Rightarrow \mathrm{Cos}B=-0.3095\\ \Rightarrow B={\cos }^{-1}(-0.3095)\\ \Rightarrow B={108.02}^{\circ }\\ \Rightarrow B={108}^{\circ }\end{array}$

Also as we know that-

  $\begin{array}{l}{a}^2=b^2+c^2-2ac\cos A\\ \Rightarrow 7^2={13}^2+9^2-2\times 13\times 9\times \cos A\\ \Rightarrow 49=169+81-234\mathrm{Cos}A\\ \Rightarrow 49=250-234\mathrm{Cos}A\\ \Rightarrow \mathrm{Cos}A=\frac{250-49}{234}\\ \Rightarrow \mathrm{Cos}A=\frac{201}{234}\\ \Rightarrow \mathrm{Cos}A=0.8589\\ \Rightarrow A={\cos }^{-1}(0.8589)\\ \Rightarrow A=30.79\\ \Rightarrow A=31\end{array}$

As we know

The sum of the angles of any triangle is ${180}^{\circ }$

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow C={180}^{\circ }-(A+B)\\ \Rightarrow C={180}^{\circ }-({31}^{\circ }+{108}^{\circ })\\ \Rightarrow C={180}^{\circ }-{139}^{\circ }\\ \Rightarrow C{=}^{}{41}^{\circ }\end{array}$

In triangle ABC

Angle $\angle A={31}^{\circ }$

Angle $\angle B={108}^{\circ }$

Angle $\angle C={41}^{\circ }$