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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.

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Answer

Let BC be the building and CD be the transmission 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. NCERT Class 10 Mathematics CBSE Solutions.

In △ABC,

tan 45° = Side opposite to angle 45°Side adjacent to angle 45°\dfrac{\text{Side opposite to angle 45°}}{\text{Side adjacent to angle 45°}}

Substituting values we get :

⇒1=BCAB⇒AB=BC=20 meters.\Rightarrow 1 = \dfrac{BC}{AB} \\[1em] \Rightarrow AB = BC = 20 \text{ meters}.

In △ABD,

tan 60° = Side opposite to angle 60°Side adjacent to angle 60°\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}

Substituting values we get :

⇒3=BDAB⇒BD=AB3=203 meters.\Rightarrow \sqrt{3} = \dfrac{BD}{AB} \\[1em] \Rightarrow BD = AB\sqrt{3} = 20\sqrt{3} \text{ meters}.

From figure,

CD = BD - BC = 203−20=20(3−1)20\sqrt{3} - 20 = 20(\sqrt{3} - 1) meters.

Hence, height of tower = 20(3−1)20(\sqrt{3} - 1) meters.

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