Chapter EP, Problem 13.5.15EP

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 13.5.15EP

The solution of the triangle is $b\approx 34.5,A\approx 55°,C\approx 80°$ .

Explanation of Solution

Given:

The given information of triangle are:

  $B=45°,a=40,c=48$

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:

  $\begin{array}{c}{b}^2={40}^2+{48}^2-2\left(40\right)\left(48\right)\cos 45°\left[\because a=40,c=48\text{ \&}\text{angle}B=45°\right]\\ b^2\approx 3904-2715.29\\ b^2\approx 1188.71\\ b\approx 34.5\end{array}$

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

  $\begin{array}{c}\frac{\sin A}{40}\approx \frac{\sin 45°}{34.5}\left[\because \frac{\sin A}{a}=\frac{\sin B}{b}\right]\\ \sin A\approx \frac{40\left(0.707\right)}{34.5}\\ \sin A\approx 0.819\\ A\approx 55°\end{array}$

The measure of the third angle is given as:

  $\begin{array}{c}m\angle C\approx 180°-\left(55°+45°\right)\\ m\angle C\approx 80°\end{array}$

Therefore, $b\approx 34.5,A\approx 55°,C\approx 80°$ .