Asurveyor 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.)

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

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3