Question

solving a real - world problem
a forest ranger in the west observation tower spots a fire 41° north of east. fifteen miles directly east, the forest ranger in the east tower spots the same fire at 56° north of west. how far away is the ranger who is closest to the fire? approximate the distance by rounding to the nearest hundredth of a mile.
9.91 mi
11.87 mi
12.53 mi
18.95 mi

Explanation:

Step1: Find the third - angle of the triangle

The two given angles of the triangle are $41^{\circ}$ and $56^{\circ}$. The sum of the interior angles of a triangle is $180^{\circ}$. Let the third - angle be $C$. Then $C=180-(41 + 56)=83^{\circ}$. The distance between the two towers is $c = 15$ miles.

Step2: Use the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. We want to find the shorter of the two distances from the towers to the fire. Let the distance from the west - tower be $a$ and from the east - tower be $b$.
We know that $\frac{a}{\sin A}=\frac{c}{\sin C}$ and $\frac{b}{\sin B}=\frac{c}{\sin C}$.
If $A = 41^{\circ}$, $B = 56^{\circ}$, and $C = 83^{\circ}$, $c = 15$ miles.
From $\frac{a}{\sin A}=\frac{c}{\sin C}$, we have $a=\frac{c\sin A}{\sin C}=\frac{15\sin41^{\circ}}{\sin83^{\circ}}$.
From $\frac{b}{\sin B}=\frac{c}{\sin C}$, we have $b=\frac{c\sin B}{\sin C}=\frac{15\sin56^{\circ}}{\sin83^{\circ}}$.
We know that $\sin41^{\circ}\approx0.6561$, $\sin56^{\circ}\approx0.8290$, $\sin83^{\circ}\approx0.9925$.
$a=\frac{15\times0.6561}{0.9925}=\frac{9.8415}{0.9925}\approx9.91$ miles.
$b=\frac{15\times0.8290}{0.9925}=\frac{12.435}{0.9925}\approx12.53$ miles.

Answer:

A. 9.91 mi