Chapter EP, Problem 13.5.3EP

To determine

To find: The given triangle can be solved by beginning with the law of Sines or law of Cosines. And hence find the solution of the given triangle

Answer to Problem 13.5.3EP

The solution of the triangle is $a\approx 18.5,m\angle B\approx 77°,m\angle C\approx 43°$ .

Explanation of Solution

Given:

The given triangle is:

As two sides and their included angle is given so the triangle will be solved by beginning with the law of Cosines.

Use the law of cosine to find the missing side length is given as:

  $a^2=b^2+c^2-2bc\cos A$

The solution of the triangle is calculated as:

  $\begin{array}{c}{a}^2={21}^2+{14}^2-2\left(21\right)\left(14\right)\cos 60°\left[\because b=21,c=14\text{ \&}\text{angle}A=60°\right]\\ a^2=637-294\\ a^2=343\\ a=18.5\end{array}$

Use the law of Sines to find the missing angle measure.

  $\begin{array}{c}\frac{\sin 60°}{18.5}\approx \frac{\sin B}{21}\left[\because \frac{\sin A}{a}=\frac{\sin B}{b}\right]\\ \sin B\approx \frac{21\left(0.86\right)}{18.5}\\ \sin B\approx 0.976\\ B\approx 77°\end{array}$

The measure of the third angle is given as:

  $\begin{array}{c}m\angle C\approx 180°-\left(60°+77°\right)\\ m\angle C\approx 43°\end{array}$

Therefore, $a\approx 18.5,m\angle B\approx 77°,m\angle C\approx 43°$ .