Chapter 16.1, Problem 2MRE

To determine

To solve: the given triangle.

Answer to Problem 2MRE

The given triangle has $b=4,\angle A=60°,\text{and}a=6.929$ .

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

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

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

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

Substitute 4for $b$ , $30°$ for $B$ and $60°$ for $A$ , and simplify

  $\begin{array}{c}a=\frac{b\sin A}{\sin B}\\ =\frac{4\sin 60°}{\sin 30°}\\ =\frac{4\times \frac{\sqrt{3}}{2}}{\frac{1}{2}}\\ =4\sqrt{3}\\ =6.929\end{array}$

Therefore, for the given triangle $b=4,\angle A=60°,\text{and}a=6.929$ .