Chapter EP, Problem 13.5.6EP

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 13.5.6EP

The solution of the triangle is $\angle A=48°,\angle B\approx 72°,m\angle C\approx 60°$ .

Explanation of Solution

Given:

The given sides of the triangle are:

  $a=19,b=24.3,c=21.8$

As given three sides of the triangle so the triangle will be solved by beginning with the law of Cosines.

Use the law of cosine to find the measure of the angle $\angle A$ is given as:

  $a^2=b^2+c^2-2bc\cos A$

The solution of the triangle is calculated as:

  $\begin{array}{c}{19}^2={24.3}^2+{21.8}^2-2\left(24.3\right)\left(21.8\right)\cos A\left[\because a=19,b=24.3,c=21.8\right]\\ 2\left(24.3\right)\left(21.8\right)\cos A=704.73\\ \cos A=0.665\\ A=48°\end{array}$

Use the law of Sines to find the missing angle $\angle B$ measure.

  $\begin{array}{c}\frac{\sin 48°}{19}\approx \frac{\sin B}{24.3}\left[\because \frac{\sin A}{a}=\frac{\sin B}{b}\right]\\ \sin B\approx \frac{24.3\left(0.743\right)}{19}\\ \sin B\approx 0.95\\ B\approx 72°\end{array}$

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

  $\begin{array}{c}m\angle C\approx 180°-\left(48°+72°\right)\\ m\angle C\approx 60°\end{array}$

Therefore, $\angle A=48°,\angle B\approx 72°,m\angle C\approx 60°$ .