Chapter EP, Problem 13.5.16EP

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 13.5.16EP

The solution of the triangle is $b\approx 13.8,A\approx 45°,C\approx 35°$ .

Explanation of Solution

Given:

The given information of triangle are:

  $B=100°,a=10,c=8$

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:

  $\begin{array}{c}{b}^2={10}^2+8^2-2\left(10\right)\left(8\right)\cos 100°\left[\because a=10,c=8\text{ \&}\text{angle}B=100°\right]\\ b^2\approx 164-\left(-27.78\right)\\ b^2\approx 191.78\\ b\approx 13.8\end{array}$

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

  $\begin{array}{c}\frac{\sin A}{10}\approx \frac{\sin 100°}{13.8}\left[\because \frac{\sin A}{a}=\frac{\sin B}{b}\right]\\ \sin A\approx \frac{10\left(0.98\right)}{13.8}\\ \sin A\approx 0.71\\ A\approx 45°\end{array}$

The measure of the third angle is given as:

  $\begin{array}{c}m\angle C\approx 180°-\left(100°+45°\right)\\ m\angle C\approx 35°\end{array}$

Therefore, $b\approx 13.8,A\approx 45°,C\approx 35°$ .