Chapter EP, Problem 13.5.17EP

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 13.5.17EP

The solution of the triangle is $\angle C\approx 75°,b\approx 4.8,a\approx 6.7$ .

Explanation of Solution

Given:

The given information of the triangle are:

  $A=40°,B=45°,c=4$

As given two angles and one side of the triangle so the triangle will be solved by beginning with the law of Sines.

Use the law of Sines to find the measure of the side $b$ is given as:

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

The measure of the third angle $\angle C$ is given as:

  $\begin{array}{c}m\angle C\approx 180°-\left(40°+45°\right)\\ m\angle C\approx 75°\end{array}$

The solution of the triangle is calculated as:

  $\begin{array}{c}\frac{\sin 40°}{a}\approx \frac{\sin 75°}{4}\\ a\approx \frac{10\left(0.64\right)}{0.96}\\ a\approx 6.7\end{array}$

Use the law of Cosines to find the missing side measure.

  $b^2=c^2+a^2-2ac\cos B$

  $\begin{array}{c}{b}^2\approx 4^2+{6.7}^2-2\left(4\right)\left(6.7\right)\cos 45°\\ b^2\approx 60.89-37.9\\ b^2\approx 22.99\\ b\approx 4.8\end{array}$

Therefore, $\angle C\approx 75°,b\approx 4.8,a\approx 6.7$ .