Chapter 9.5, Problem 41E

Solution

To find: the area of the $△ABC$ .

  $\left.\text{A}\right){\text{274 units}}^2$

Given information:

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

  $B=52°,a=29,c=24$

The given options are:

  $274{\text{ units}}^2$

  $348{\text{ units}}^2$

  $548{\text{ units}}^2$

  $696{\text{ units}}^2$

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:

  

  $Area=\frac{1}{2}bc\sin A$

  $Area=\frac{1}{2}ac\sin B$

  $Area=\frac{1}{2}ab\sin C$

Explanation:

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

  $\begin{array}{c}Area=\frac{1}{2}ac\sin B\\ =\frac{1}{2}\cdot 29\cdot 24\cdot \sin 52°\\ \approx 274.22\\ \approx 274\text{ square units}\end{array}$

Option (A) is the correct choice.