Chapter 9.5, Problem 7GP

Solution

To find: the area of the $△ABC$ .

  $Area\approx 50.35\text{ square units}$

Given information:

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

  $a=10,b=14,C=46°$

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:

  $\begin{array}{c}Area=\frac{1}{2}ab\sin C\\ =\frac{1}{2}\cdot 10\cdot 14\cdot \sin 46°\\ \approx 50.35\text{ square units}\end{array}$