Heights and Distances Practice Test 1

Question : 4 of 40

Marks: +1, -0

ABC is a triangular park with AB = AC = 100 metres. A clock tower is situated at the mid-point of BC. The angles of elevation of the top of the tower of A and B are

\cot^{–1} (3.2)

and

\cosec^{–1} (2.6)

respectively. The height of the tower in metres is:

$25/2$
25
50
None of these

Validate

Solution:

Given

α = cot^{–1} (3.2)

β = cosec^{–1} (2.6)

In ΔPAD, AD = h cot α,
In ΔPBD, BD = h cot β
In ΔABD,

AB^2 = AD^2 + BD^2

= h^2 (\cot^2 α + \cot^2 β)

⇒ 100^2 = h^2 \{\cot^2α + (\cosec^2 α - 1)\}

⇒ 100^2 = h^2 [(3.2)^2 + (2.6)^2 - 1] = 16h^2

⇒ h = 25 m.

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