Answer:
41.7feet
Step-by-step explanation:
From the question we are given the following
angle of depression = 50°
Distance of the pole from the base of the feet = 35feet (Adjacent)
Required
height of the school (opposite)
Using the SOH CAH TOA identity
Tan theta = opp/adj
Tan 50 = H/35
H = 35tan 50
H = 35(1.1918)
H = 41.7feet
Hence the height of the school is 41.7feet
