Chapter 14.4, Problem 5E

Solution

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

Two sets measurements of angles $\angle B$ and $\angle C$ are: $m\angle B=44.16°\text{, }m\angle C={112.16}^{\text{o}}$ and

  $m\angle B=135.84°,m\angle C={20.48}^{\text{o}}$

Given data: The measure of angle $m\angle A=23.68°$ , $a=9.8$ , and $b=17$ .

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 $m\angle A={23.68}^{\text{o}}$ , $a=9.8$ , and $b=17$ .

Substitute ${23.68}^{\text{o}}$ for $\angle A$ , $9.8$ for ‘ a’ and $17$ for b in the result (1),

  $\frac{\sin 23.68°}{9.8}=\frac{\sin \angle B}{17}=\frac{\sin \angle C}{c}$

Taking first two ratios,

  $\frac{\sin 23.68°}{9.8}=\frac{\sin \angle B}{17}$

  $\begin{array}{c}\sin \angle B=\frac{17\sin 23.68°}{9.8}\\ \angle B={\sin }^{-1}\left(\frac{17\sin 23.68°}{9.8}\right)\\ ={\sin }^{-1}\left(0.6967\right)\\ =44.16°\end{array}$

Since sine is positive in second quadrant so another value of $\angle B$ is $180°-44.16°$ or $135.84°$

Thus, in triangle ABC , $m\angle A=23.68°$ , $m\angle B={44.16}^{\text{o}}$ , therefore, from result (2),

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

Substitute $23.68°$ for $m\angle A$ and ${44.16}^{\text{o}}$ for $m\angle B$

  $\begin{array}{c}m\angle C={180}^{\text{o}}-\left(23.68°+{44.16}^{\text{o}}\right)\\ ={112.16}^{\text{o}}\end{array}$

Substitute $23.68°$ for $m\angle A$ and $135.84°$ for $m\angle B$

  $\begin{array}{c}m\angle C={180}^{\text{o}}-\left(23.68°+135.84°\right)\\ ={20.48}^{\text{o}}\end{array}$

Thus, two sets measurements of angles $\angle B$ and $\angle C$ are: $m\angle B=44.16°\text{, }m\angle C={112.16}^{\text{o}}$ and

  $m\angle B=135.84°,m\angle C={20.48}^{\text{o}}$ .