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