Chapter 14.4, Problem 2E

Solution

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

The set of two triangles formed by given data is

  $\left.\begin{array}{c}\angle A={16}^{\text{ο}},\angle B={\mathrm{59.5.}}^{\text{ο}},\angle C={104.5}^{\text{ο}}\\ a=12\text{cm},b=37.5\text{cm},\text{}\text{}\text{}\text{}\text{}c=42\text{cm}\end{array}\right\}$ , $\begin{array}{c}\angle A={16}^{\text{ο}},\angle B={121.5}^{\text{ο}},\angle C={42.5}^{\text{ο}}\\ a=12\text{cm},b=37.5\text{cm},\text{}\text{}\text{}\text{}\text{}c=29.4\text{cm}\end{array}$

Given data: The measure of angle $m\angle A={16}^{\text{o}}$ , $a=12$ , and $b=37.5$

Method/Formula used:

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

  

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=K\left(\text{constant}\right)$ … (1)

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

OR $\angle A+\angle B+\angle C={180}^{\text{o}}$ … (2)

Calculations:

The measure of angle $m\angle A={16}^{\text{o}}$ , $a=12$ , and $b=37.5$

Use Sin Law for given data of triangle ABC,

  $K=\frac{\sin \angle A}{a}$

Substitute ${16}^{\text{o}}$ for $\angle A$ , 12 for a

  $\begin{array}{c}K=\frac{\sin \angle A}{a}\leftarrow \angle A={16}^{\text{ο}},a=12\\ =\frac{\sin {16}^{\text{ο}}}{12}\\ =0.023\end{array}$

Therefore,

  $\begin{array}{c}\frac{\sin \angle B}{b}=0.023\leftarrow b=37.5\\ \sin \angle B=0.023\times 37.5\\ \angle B={\sin }^{-1}\left(0.8614\right)\\ \angle B={59.5}^{\text{ο}}\end{array}$

Since, sine functions has two positive values when angle is less than ${180}^{\text{ο}}$ , if one is $\angle B$ then other is ${180}^{\text{ο}}-\angle B$ So, there are two values of angle $\angle B$ , ${59.5}^{\text{ο}}$ and ${180}^{\text{ο}}-{59.5}^{\text{ο}}={121.5}^{\text{ο}}$

Substitute ${16}^{\text{ο}}$ for A and ${59.5}^{\text{o}}$ for $\angle B$ , in result (2),

  $\begin{array}{c}{16}^{\text{o}}+{59.5}^{\text{o}}+\angle C={180}^{\text{o}}\\ \angle C={180}^{\text{o}}-\left({16}^{\text{o}}+{59.5}^{\text{o}}\right)\\ ={104.5}^{\text{o}}\end{array}$

Substitute ${16}^{\text{ο}}$ for A and ${121.5}^{\text{ο}}$ for $\angle B$ , in result (2),

  $\begin{array}{c}{16}^{\text{o}}+{121.5}^{\text{ο}}+\angle C={180}^{\text{o}}\\ \angle C={180}^{\text{o}}-\left({16}^{\text{o}}+{121.5}^{\text{ο}}\right)\\ ={42.5}^{\text{o}}\end{array}$

Sine in a triangle only one angle can be obtuse, therefore, the triangle can have angles as:

  $\angle A={16}^{\text{o}},\angle B={59.5}^{\text{o}},\angle C={104.5}^{\text{o}}$

Therefore,

  $\begin{array}{c}c=\frac{\sin {104}^{\text{ο}}}{0.023}\\ =42\end{array}$

OR

  $\angle A={16}^{\text{o}},\angle B={121.5}^{\text{o}},\angle C={42.5}^{\text{o}}$

Therefore,

  $\begin{array}{c}c=\frac{\sin {42.5}^{\text{ο}}}{0.023}\\ \approx 29.4\end{array}$

Therefore, the set of two triangles formed by given data is

  $\begin{array}{c}\left.\begin{array}{c}\angle A={16}^{\text{ο}},\angle B={\mathrm{59.5.}}^{\text{ο}},\angle C={104.5}^{\text{ο}}\\ a=12\text{cm},b=37.5\text{cm},\text{}\text{}\text{}\text{}\text{}c=42\text{cm}\end{array}\right\}\\ \left.\begin{array}{c}\angle A={16}^{\text{ο}},\angle B={121.5}^{\text{ο}},\angle C={42.5}^{\text{ο}}\\ a=12\text{cm},b=37.5\text{cm},\text{}\text{}\text{}\text{}\text{}c=29.4\text{cm}\end{array}\right\}\end{array}$