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The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is $60 °$ . At a point, Y. 40 m vertically above X, the angle of elevation is $45 °$ . Find the height of the tower PQ and the distance XQ.

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195.64 m and 109.3 m

94.64 m and 100.3 m

84.64 m and 109.3 m

94.64 m and 109.3 m

answer is C.

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Detailed Solution

Given that the angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is

60 °

.
We have to find the height of the tower PQ and the distance XQ.
The value for the trigonometric identities can be evaluated by using the right-angled triangle the value for can be expressed as

tan θ = opposite adjacent

The required figure geometry is shown in figure below,
Question Image

Then by using the figure geometry, in triangle YRQ

tan θ = Q R Y R ⇒ tan 45 ° = x γ R ⇒ 1 = x γ R ⇒ x = Y R

Since,

Y R = X P .

So,

X P = X

Similarly, in triangle XPQ,

tan θ = P Q P X ⇒ tan 60 ° = x + 40 x ⇒ 3 x = x + 40 ⇒ x = 40 3 − 1

Thus, on further solving the above expression,

x = 40 3 − 1 ⋅ 3 + 1 3 + 1 ⇒ x = 20 (3 + 1) ⇒ x = 20 (173 + 1) ⇒ x = 54.64 m

P Q = x + 40 ⇒ P Q = 54.64 + 40 ⇒ P Q = 94.64 m Similarly, in the triangle QPX

sin 60 ° = P Q X Q ⇒ 32 = 94.64 X Q ⇒ X Q = 94.64 (2) 3 ∴ X Q = 109 .3m

The height of the tower PQ and the distance XQ 94.64 m and 109.3 m respectively.
Therefore, the correct option is 3.

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