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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 ?
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.
By: Amit Kumar ProfileResourcesReport error
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