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Question

A window in a building is at a height of 10 m from the ground. The angle of depression of a point P on the ground from the window is 30°. The angle of elevation of the top of the building from the point P is 60°. Find the height of the building.

Or, in a violent storm, a tree is bent by the wind. The top of the tree meets the ground at an angle of 30°, at a distance of 30 metres from the root. At what height from the bottom did the tree get bent? What was the original height of the tree?

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Solution

Given, height of the window from the ground DB = 10 m
Let H be the height of the building.
Let O be the point of observation. Then,

DB = 10 m, DPB = 30° and APB = 60°
BP = x m
tan30°=10x
13=10xx=103m ...............(i)
Again,
tan60°=Hx
3=Hx
H=3x
H=3×103m=30 m
Hence, required height of the building is 30 m.

OR
Let AB be the original height of the tree, which gets bent at a point C. After being bent, let the part CB take the position CD, meeting the ground at D. Then AD = 30 m, ∠ADC = 30°, ∠DAC = 90° and CD = CB

Let AC = x metres and CD = CB = y metres
From right-angled ΔDAC, we have:
ACAD=tan30°=13
x30=13
x=30×13×33=103m
Also, from right-angled ΔDAC, we have:
DCAD=sec30°=23
y30=23
y=30×23×33=203m
Thus AB = (x + y) metres = 103+203m=303m
Hence, the tree got bent at a height of 103m and the original height of the tree was 303m.

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