Question

Two forest rangers are standing on a road that runs east-west. One sights a forest fire in the direction 31° east of

north. The other sights the same fire 57° west of north.
Find the distance from each ranger to the forest fire and
the shortest distance from the fire to the road.

Answer 1

Answer: The forest ranger closest to the fire is at a distance of 9.92 miles (nearest hundredth).

Draw a horizontal line AB of length 15 miles to represent the positions of the two forest rangers.

Point A is to the left and point B is to the right.

Complete triangle ABC so that m∠A = 41° and m∠B = 56°.

The point C represents the location of the fire.

Let x = length of BC. It is the shortest distance from the forest ranger B to the fire because BC is opposite the smaller angle of 41°.

Because the sum of angles in a triangle is 180°,

m∠C = 180° - (41° + 56°) = 83°.

From the Law of Sines,

x/sin(41) = 15/sin(83)

x = 15*[sin(41)/sin(83)] = 9.915 mi

Explanation:

Draw a horizontal line AB of length 15 miles to represent the positions of the two forest rangers.

Point A is to the left and point B is to the right.

Complete triangle ABC so that m∠A = 41° and m∠B = 56°.

The point C represents the location of the fire.

Let x = length of BC. It is the shortest distance from the forest ranger B to the fire because BC is opposite the smaller angle of 41°.

Because the sum of angles in a triangle is 180°,

m∠C = 180° - (41° + 56°) = 83°.

From the Law of Sines,

x/sin(41) = 15/sin(83)

x = 15*[sin(41)/sin(83)] = 9.915 mi