Chapter 9, Problem 22T

Solution

To find: the area of the $△ABC$ .

  $\text{142}{\text{.23 units}}^2$

Given information:

The side lengths and the included angle of the triangle is:

  $A=81°,b=16,c=18$

Definition Used:

Area of a triangle

The area of any triangle is given by one half the product of the lengths of two sides times the sine of their included angle. For $△ABC$

shown, there are three ways to calculate the area:

  

  $\begin{array}{l}Area=\frac{1}{2}bc\sin A\\ Area=\frac{1}{2}ac\sin B\\ Area=\frac{1}{2}ab\sin C\end{array}$

Explanation:

Using the above mentioned formula, the area of the given triangle is:

  $\begin{array}{c}Area=\frac{1}{2}bc\sin A\\ =\frac{1}{2}\cdot 16\cdot 18\cdot \sin 81°\\ \approx 142.23\text{ square units}\end{array}$