Chapter 12.9, Problem 13WE

To determine

Find the largest angle of a triangle which has area of 48 square centimeter and shorter sides have length 9cm and 12cm respectively.

Answer to Problem 13WE

The largest angle is $117.3°$ .

Explanation of Solution

Given information: The area of triangle is 48 square centimeter and shorter sides are 9cm and 12cm respectively.

Calculation: Since shorter sides are 9cm and 12 cm respectively, therefore they will include the largest possible angle in between them because largest side will be opposite to this largest angle. Therefore if $\left(\theta \right)$ is the angle between these sides then by formula of area $\left(A\right)$ of triangle,

  $\begin{array}{c}\text{A}=\frac{1}{2}ab\sin \left(\theta \right)\\ 48=\frac{1}{2}\left(9\right)\left(12\right)\left(\sin \left(\theta \right)\right)\\ \text{or, }\sin \theta =\frac{8}{9}\\ \theta ={\sin }^{-1}\frac{8}{9}\\ \theta =62.7°\end{array}$

Now the largest angle can be obtained by subtracting $\left(\theta \right)$ from $180°$ .

  $\begin{array}{c}\text{Largest angle}=180°-62.7°\\ =117.3°\end{array}$

Hence the largest angle is $117.3°$ .