Question

A surveyor sights the top of a building with a handheld range finder. the top of the building is 148 feet 2 inches away. the angle of elevation is 56º. find the distance from the surveyor to the building in feet and inches. (sketch and label the situation first.)

Answer 1

see the attached figure to better understand the problem

we know that

in the right triangle ABC

cos 56°=AC/AB

where

AC is the adjacent side to angle 56 degrees------> the distance from the surveyor to the building

AB is the hypotenuse-----> 148 ft 2 in

56 degrees------> is the angle of elevation

so

cos 56°=AC/AB---------> solve for AC

AC=AB*cos 56°

AB=148 ft 2 in

convert 2 in to ft

1 ft -----> 12 in

x ft------> 2 in

x=2/12-----> x=0.17 ft

AB=148 ft 2 in-----> 148 ft+0.17 ft------> AB=148.17 ft

AC=AB*cos 56°----> AC=148.17*cos 56°------> AC=82.86 ft

convert 0.86 ft to in

0.86 ft=0.86*12-----> 10.32 in

distance AB=82 ft 10 in

the answer is

the distance from the surveyor to the building is 82 ft 10 in