# The angle of elev

In the figure, let AB be the tower and C be the point on the ground from which the angle of elevation of the top of the tower AB is 60°. Join A and C and B and C. Then we get a right-angled triangle ABC with right angle at B and ACB = 60°. Let D be the point 10 m vertically above C. Then, CD = 10 m. Given that the angle of elevation of the top of the tower from the point D is 30°. Join D and A. Also draw a line DE from D to AB, parallel to BC. Then we get a right-angled triangle AED with right angle at E and ADE = 30°. We are to find the height of the tower, that is AB.

Clearly, ED = BC. Let AE = x

In ∆AED,

or,

or,

Again, in ∆ABC,

or,

or,

or,

3x = x + 10"

or,

x = 5m

Height of the tower = AB = AE + EB = 5m + 10m = 15m.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses