Chapter 16.1, Problem 6MRE

To determine

To find: the area of the given triangle.

Answer to Problem 6MRE

Area of the given triangle is $13.856$ square units.

Explanation of Solution

Given information:

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

  $c=8,\angle C=90°,\angle B=30°$

Calculation:

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

Subtract $90°$ and $30°$ from $180°$ to find $A$ ,

  $180°-90°-30°=60°$

Next, to find the sides.

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

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

Substitute 8 for $c$ , $30°$ for $B$ and $90°$ for $C$ , and simplify

  $\begin{array}{c}b=\frac{c\sin B}{\sin C}\\ =\frac{8\sin 30°}{\sin 90°}\\ =\frac{8\times \frac{1}{2}}{1}\\ =4\end{array}$

Use the following formula to find area of the triangle.

  $A=\frac{1}{2}bc\sin A$

Substitute 4for $b$ , 8 for $c$ and $60°$ for $A$ , and simplify

  $\begin{array}{c}A=\frac{1}{2}bc\sin A\\ =\frac{1}{2}\left(4\right)\left(8\right)\sin 60°\\ =16\times \frac{\sqrt{3}}{2}\\ \approx 13.856\end{array}$

Therefore, area of given triangle is $13.856$ square units.