Chapter EP, Problem 13.5.18EP

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 13.5.18EP

The solution of the triangle is $a=35.6,m\angle B\approx 73°,m\angle C\approx 87°$ .

Explanation of Solution

Given:

The given information of the triangle are:

  $A=20°,b=100,c=84$

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:

  $\begin{array}{c}{a}^2\approx {100}^2+{84}^2-2\left(100\right)\left(84\right)\cos 20°\left[\because b=100,c=84\text{ \&}\text{angle}A=20°\right]\\ a^2\approx 17056-15786.8\\ a^2\approx 1269.2\\ a\approx 35.6\end{array}$

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

  $\begin{array}{c}\frac{\sin 20°}{35.6}\approx \frac{\sin B}{100}\left[\because \frac{\sin A}{a}=\frac{\sin B}{b}\right]\\ \sin B\approx \frac{100\left(0.34\right)}{35.6}\\ \sin B\approx 0.955\\ B\approx 73°\end{array}$

The measure of the third angle is given as:

  $\begin{array}{c}m\angle C\approx 180°-\left(20°+73°\right)\\ m\angle C\approx 87°\end{array}$

Therefore, $a=35.6,m\angle B\approx 73°,m\angle C\approx 87°$ .