Chapter 13.4, Problem 39PPS

To determine

To find:How far apart are sirens B and C.

Answer to Problem 39PPS

Distance between sirens B and C is $15.8$ miles

Explanation of Solution

Given information:

  $\begin{array}{l}A={112}^0\\ B={40}^0\\ c=8\end{array}$

Given

  $\begin{array}{l}A={112}^0\\ B={40}^0\\ c=8\end{array}$

We know

Sum of angles of a triangle is ${180}^0$

  $\begin{array}{l}\Rightarrow A+B+C={180}^0\\ \Rightarrow {112}^0+{40}^0+C={180}^0\\ \Rightarrow {152}^0+C={180}^0\\ \Rightarrow C={180}^0-{152}^0\\ \Rightarrow C={28}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin C}{c}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin {28}^0}{8}=\frac{\sin {112}^0}{a}\\ \Rightarrow a=\frac{8\sin {112}^0}{\sin {28}^0}\\ \Rightarrow a=\frac{8\left(0.927\right)}{0.47}\\ \Rightarrow a=\frac{7.416}{0.47}\\ \Rightarrow a=15.8\end{array}$

Therefore,

Distance between sirens B and C is $15.8$ miles