Q. 205.0( 4 Votes )
The angle of depression from the top of a tower of a point A on the ground is 30°. On moving a distance of 20 meters from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower from the point B is 60°. Find the height of the tower and its distance from the point A.
Let DC be the tower. Given that the angle of depression of the point A on the ground from the top of the tower DC is 30°. Join C and A. Now draw a line DE parallel to CA. Also join A and D. Then, ∠ EDA = 30°. We get a right-angled triangle ACD with right angle at C and ∠DAC = 30°. If we move 20 m from A to B towards the foot of the tower C, then the angle of depression changes to 60°. Then, AB = 20 m. Join D and B. Then we get a right-angled triangle ∆DCB with right angle at C.
We are to find the height of the tower that is DC and its distance from A, that is, AC.
Let DC = x. In ∆DCB,
The height of the tower = DC = x = 17.32m.
We have, BC = x/√3 = 10m. So, a distance of A from the tower = AC = 20 + 10 = 30m.
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