Question:
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Updated On: Jan 13, 2026
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Solution and Explanation
Let BC be the building, AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured.
In ∆BCD,
\(\frac{BC}{CD} = tan45° \)
\(\frac{20}{ CD} =1\)
\(CD = 20m\)
In ∆ACD,
\(\frac{AC}{ CD }= tan 60°\)
\(\frac{AB + BC}{ CD} = \sqrt3\)
\(\frac{AB + 20}{ 20} = \sqrt3\)
\(AB = (20\sqrt3 -20)\, m\)
\(AB = 20(\sqrt3 -1)\,m\)
Therefore, the height of the transmission tower is \(20(\sqrt3 -1)\,m\).
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