Chapter EP, Problem 13.5.10EP

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 13.5.10EP

The solution of the triangle is $c\approx 5.4,B\approx 44°,m\angle A\approx 111°$ .

Explanation of Solution

Given:

The given information of the triangle is:

  $b=9,a=12,C=25°$

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:

  $c^2=a^2+b^2-2ab\cos C$

The solution of the triangle is calculated as:

  $\begin{array}{c}{c}^2=9^2+{12}^2-2\left(9\right)\left(12\right)\cos 25°\left[\because b=9,a=12\text{ \&}\text{angle}C=25°\right]\\ c^2\approx 225-195.8\\ c^2\approx 29.2\\ c\approx 5.4\end{array}$

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

  $\begin{array}{c}\frac{\sin 25°}{5.4}\approx \frac{\sin B}{9}\left[\because \frac{\sin C}{c}=\frac{\sin B}{b}\right]\\ \sin B\approx \frac{9\left(0.42\right)}{5.4}\\ \sin B\approx 0.7\\ B\approx 44°\end{array}$

The measure of the third angle is given as:

  $\begin{array}{c}m\angle A\approx 180°-\left(25°+44°\right)\\ m\angle A\approx 111°\end{array}$

Therefore, $c\approx 5.4,B\approx 44°,m\angle A\approx 111°$ .