Chapter 13.1, Problem 18WE

To determine

To find: The all to three significant digits and angles.

Answer to Problem 18WE

The angle $\angle A\approx {100}^o$ , $\angle B\approx {38}^o$ , and $c\approx 5.44$ .

Explanation of Solution

Given information:

The given sides of triangle are $a=8$ , $b=5$ , and angle $\angle C={42}^o$ .

Calculation:

Consider the values summarize in the figure.

  

Use the law of cosine to find $c$ .

  $c^2=a^2+b^2-2ab\cos C$

Substitute the values of $a,b$ , and angle $C$ .

  $\begin{array}{c}{c}^2=8^2+5^2-2\left(8\right)\left(5\right)\cos {42}^o\\ =64+25-80\cos {42}^o\\ =89-59.45\\ \approx 29.54\end{array}$

Take the square root on the both sides.

  $\begin{array}{c}\sqrt{c^2}\approx \sqrt{29.54}\\ c\approx 5.44\end{array}$

Now, use the law sine to find the measure of angle $C$ .

  $\frac{\sin B}{b}=\frac{\sin C}{c}$

Substitute the value of $b,c$ , to get the $C$ .

  $\begin{array}{c}\frac{\sin B}{5}=\frac{\sin {42}^o}{5.44}\\ \sin B=5\left(\frac{\sin {42}^o}{5.44}\right)\\ \approx 5\left(0.123\right)\\ \approx 0.615\end{array}$

Use the inverse sine function.

  $\begin{array}{c}B={\sin }^{-1}0.615\\ \approx {38}^o\end{array}$

Since the sum of all three angles is ${180}^o$ .

  $\begin{array}{c}\angle A+\angle B+\angle C={180}^o\\ \angle A+{38}^o+{42}^o={180}^o\\ \angle A={180}^o-{80}^o\\ ={100}^o\end{array}$

Therefore, the angle $\angle A\approx {100}^o$ , $\angle B\approx {38}^o$ , and $c\approx 5.44$ .