Chapter 9.5, Problem 40E

Solution

To find: the area of the $△ABC$ .

  $Area\approx 147.3\text{ square units}$

Given information:

The image of $△ABC$:

  

Definition Used:

Law of Sines

The law of sines can be written in either of the following forms for $△ABC$ with sides of length a , b and c .

  

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

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:

Comparing the given triangle with the following figure:

  

  $A=66°,a=20,c=16$

Now, by the law of Sines,

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \frac{\sin 66°}{20}=\frac{\sin C}{16}\\ \sin C=\frac{16\sin 66°}{20}\\ C={\sin }^{-1}\left(\frac{16\sin 66°}{20}\right)\\ C\approx 47°\end{array}$

It is known that the sum of all the angles of a triangle is $180°$ .

Using this knowledge to find angle A,

  $\begin{array}{c}A+B+C=180°\\ 66°+B+47°\approx 180°\\ B\approx 180°-66°-47°\\ B\approx 67°\end{array}$

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

  $\begin{array}{c}Area=\frac{1}{2}ac\sin B\\ =\frac{1}{2}\cdot 20\cdot 16\cdot \sin 67°\\ \approx 147.3\text{ square units}\end{array}$