Question

The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower ?
A) 42.2 mts
B) 33.45 mts
C) 66.6 mts
D) 58.78 mts

Answer 1

The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower ?

A) 42.2 mtsB) 33.45 mtsC) 66.6 mtsD) 58.78 mts

Answer:   C) 66.6 mts 

Explanation:



From above diagram
AC represents the hill and DE represents the tower

Given that AC = 100 m

angleXAD = angleADB = 30° (∵ AX || BD ) 
angleXAE = angleAEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)

tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)
∵ BD = CE and Substitute the value of CE from equation 1 
100/√3 = 100−h(√3) => h = 66.66 mts

The height of the tower = 66.66 mts.

Answer 2

AC represents the hill and DE represents the tower

Given that AC = 100 m

angleXAD = angleADB = 30° (∵ AX || BD )

angleXAE = angleAEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)

tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)

∵ BD = CE and Substitute the value of CE from equation 1

100/√3 = 100−h(√3) => h = 66.66 mts

The height of the tower = 66.66 mts.