Chapter 14.5, Problem 40PPSE

Solution

To find: the value of $m\angle B$ to the nearest tenth.

  $m\angle B\approx {62.7}^{\circ }$ .

Given:

In the triangle ABC, $m\angle A={53}^{\circ },c=7$ cm, and $b=37$ cm.

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 ABC, it follows that

  $\begin{array}{c}{a}^2=b^2+c^2-2bc\cos A\\ a^2={\left(37\right)}^2+7^2-2\left(37\right)\left(7\right)\cos {53}^{\circ }\\ a^2=1418-311.7401\\ a^2=1106.2598\\ a=33.2604\\ a\approx 33.3\end{array}$

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

  $\begin{array}{c}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \frac{\sin {53}^{\circ }}{33.3}=\frac{\sin B}{37}\\ \sin B=\frac{37\times \sin {53}^{\circ }}{33.3}\\ \sin B=0.8884\\ m\angle B={\sin }^{-1}\left(0.8884\right)\\ m\angle B={62.67}^{\circ }\\ m\angle B\approx {62.7}^{\circ }\end{array}$