Chapter 14.4, Problem 4E

Solution

To find: The measures of angles $\angle B$ and $\angle C$ of a triangle ABC .

The measurements of angles $\angle B$ and $\angle C$ are $m\angle B=17.7°$ and $m\angle C={50.3}^{\text{o}}$ .

Given data: The measure of angle $m\angle A=112°$ , $a=16.5$ , and $b=5.4$ .

Method/Formula used:

The Sine Law of triangle ABC shown in Fig. 1 is

  

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ … (1)

The sum of angles of triangle $m\angle A+m\angle B+m\angle C={180}^{\text{o}}$ … (2)

Calculations:

The measure of angle $m\angle A={112}^{\text{o}}$ , $a=16.5$ , and $b=5.4$ .

Substitute ${112}^{\text{o}}$ for $\angle A$ , $16.5$ for ‘ a’ and $5.4$ for b in the result (1),

  $\frac{\sin 112°}{16.5}=\frac{\sin \angle B}{5.4}=\frac{\sin \angle C}{c}$

Taking first two ratios,

  $\frac{\sin 112°}{16.5}=\frac{\sin \angle B}{5.4}$

  $\begin{array}{c}\sin \angle B=\frac{5.4\sin 112°}{16.5}\\ \angle B={\sin }^{-1}\left(\frac{5.4\sin 112°}{16.5}\right)\\ =17.7°\end{array}$

Since sine is positive in second quadrant so another value of $\angle B$ is $180°-17.7°$ or $162.3°$

Thus, in triangle ABC , $m\angle A={112}^{\text{o}}$ , $m\angle B={17.7}^{\text{o}}$ , therefore, from result (2),

  $m\angle C={180}^{\text{o}}-\left(m\angle A+m\angle B\right)$

Substitute ${112}^{\text{o}}$ for $m\angle A$ and ${17.7}^{\text{o}}$ for $m\angle B$

  $\begin{array}{c}m\angle C={180}^{\text{o}}-\left(112°+17.7°\right)\\ ={50.3}^{\text{o}}\end{array}$

Since in a triangle only one angle can be obtuse, therefore, $\angle B\ne 162.3°$ .

Therefore, the measurements of angles $\angle B$ and $\angle C$ are $m\angle B=17.7°$ and $m\angle C={50.3}^{\text{o}}$ .