Chapter 14.5, Problem 19PPSE

Solution

To find: the value of $m\angle S$ .

  $m\angle S\approx {26.96}^{\circ }$ .

Given:

In triangle RST, $r=17$ cm, $s=12$ cm, and $m\angle T={13}^{\circ }$ .

Concept used:

Law of Cosines:

In a triangle ABC, let $a,b$ , and $c$ represents the length of the sides opposite $\angle A,\angle B$ , and $\angle C$ , respectively. Then,

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ b^2=a^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C\end{array}$

Law of Sines:

In any triangle ABC with side lengths $a,b$ , and $c$ ,

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ .

Calculation:

By using the Law of Cosines in the given triangle, it follows that

  $\begin{array}{c}{t}^2=r^2+s^2-2rs\cos T\\ t^2={17}^2+{12}^2-2\left(17\right)\left(12\right)\cos {13}^{\circ }\\ t^2=433-397.5429\\ t^2=35.4570\\ t=5.9545\\ t\approx 5.95\end{array}$

Now by using the Law of Sines in the given triangle, it follows that

  $\begin{array}{c}\frac{\sin S}{s}=\frac{\sin T}{t}\\ \frac{\sin S}{12}=\frac{\sin {13}^{\circ }}{5.95}\\ \sin S=\frac{12\times \sin {13}^{\circ }}{5.95}\\ \sin S=0.4533\\ m\angle S={\sin }^{-1}\left(0.4533\right)\\ m\angle S\approx {26.96}^{\circ }\end{array}$