Chapter 16.1, Problem 8MRE

To determine

To find: the area of the given triangle.

Answer to Problem 8MRE

Area of the given triangle is $194.015$ square units.

Explanation of Solution

Given information:

Given lengths to three significant digits and angle measures to the nearest tenth of a degree.

  $a=24,b=17,\angle A=72°$

Calculation:

Let us find angle $B$ using the law of sines.

  $\frac{\sin B}{b}=\frac{\sin A}{a}\text{or}\sin B=\frac{b\sin A}{a}$

Substitute $24$ for $a$ , 17 for $b$ and $72°$ for $A$ , and simplify

  $\begin{array}{c}\sin B=\frac{17\sin 72°}{24}\\ =0.673\end{array}$

Use the inverse sine function to find $B$ .

  $\begin{array}{c}B={\sin }^{-1}\left(0.673\right)\\ =42°\end{array}$

The sum of all the angles on a triangle is $180°$ .

Subtract $72°$ and $42°$ from $180°$ to find $C$ ,

  $180°-72°-42°=66°$

Let us find side $c$ using the law of sines.

  $\frac{\sin B}{b}=\frac{\sin C}{c}\text{or}c=\frac{b\sin C}{\sin B}$

Substitute $42°$ for $B$ , 17 for $b$ and $66°$ for $C$ , and simplify

  $\begin{array}{c}c=\frac{17\sin 66°}{\sin 42°}\\ =23.006\end{array}$

Use the following formula to find area of the triangle.

  $A=\frac{1}{2}bc\sin A$

Substitute 17for $b$ , 23.006 for $c$ and $72°$ for $A$ , and simplify

  $\begin{array}{c}A=\frac{1}{2}bc\sin A\\ =\frac{1}{2}\left(17\right)\left(23.006\right)\sin 72°\\ \approx 194.015\end{array}$

Therefore, area of given triangle is $194.015$ square units.