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 18.95 mi

Explanation:

Step1: Find the third angle of the triangle

The sum of angles in a triangle is \(180^\circ\). Let the angle at the fire be \(C\). So \(C = 180^\circ - 41^\circ - 56^\circ = 83^\circ\).

Step2: Apply the Law of Sines

Let \(a\) be the distance from the east tower to the fire, \(b\) be the distance from the west tower to the fire, and \(c = 15\) miles (distance between the two towers). By the Law of Sines, \(\frac{a}{\sin 41^\circ}=\frac{b}{\sin 56^\circ}=\frac{c}{\sin 83^\circ}\).
First, find \(a\) (distance from east tower to fire): \(\frac{a}{\sin 41^\circ}=\frac{15}{\sin 83^\circ}\), so \(a=\frac{15\sin 41^\circ}{\sin 83^\circ}\).
Calculate \(\sin 41^\circ\approx0.6561\), \(\sin 83^\circ\approx0.9925\). Then \(a=\frac{15\times0.6561}{0.9925}\approx\frac{9.8415}{0.9925}\approx9.91\) miles.
Find \(b\) (distance from west tower to fire): \(\frac{b}{\sin 56^\circ}=\frac{15}{\sin 83^\circ}\), \(\sin 56^\circ\approx0.8290\), so \(b = \frac{15\times0.8290}{0.9925}\approx\frac{12.435}{0.9925}\approx12.53\) miles.
Compare \(a\) and \(b\), \(9.91<12.53\), so the closest ranger (east tower) is about 9.91 miles from the fire.

Answer:

9.91 mi