Aforest 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

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