Chapter 12.9, Problem 14WE

To determine

Find the longest side of a triangle which has area of 75 square centimeter and three angles are $25°,45°\text{ and }110°$ respectively.

Answer to Problem 14WE

The longest side is $23.1\text{ cm}$ .

Explanation of Solution

Given information: The area of triangle is 75 square centimeter and three angles are $25°,45°\text{ and }110°$ respectively.

Calculation: Since the three angles are $25°,45°\text{ and }110°$ respectively, therefore longest side will be opposite to largest angle. Now by formula of area $\left(A\right)$ of triangle,

  $\begin{array}{c}A=\frac{1}{2}{\left(c\right)}^2\frac{\sin A\sin B}{\sin C}\\ 75=\frac{1}{2}{c}^2\frac{\sin \left(22\right)\sin \left(45\right)}{\sin 110}\\ \text{or, }{c}^2=\frac{150\sin \left(110\right)}{\sin \left(22\right)\sin \left(45\right)}\\ c=\sqrt{\frac{150\sin \left(110\right)}{\sin \left(22\right)\sin \left(45\right)}}\\ c=23.1\text{ cm}\end{array}$

Hence the longest side is $23.1\text{ cm}$ .