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From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

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Answer

Let AB be the building and CD be the cable tower.

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. NCERT Class 10 Mathematics CBSE Solutions.

Given,

Angle of depression of foot of tower from top of building is 45°.

∴ ∠EAC = 45°

From figure,

∠ACB = ∠EAC = 45° [Alternate angles are equal].

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=ABBC⇒BC=AB=7 m.\Rightarrow 1 = \dfrac{AB}{BC} \\[1em] \Rightarrow BC = AB = 7 \text{ m}.

From figure,

AE = BC = 7 m.

In △ADE,

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=DEAE⇒3=DE7⇒DE=73 m.\Rightarrow \sqrt{3} = \dfrac{DE}{AE} \\[1em] \Rightarrow \sqrt{3} = \dfrac{DE}{7} \\[1em] \Rightarrow DE = 7\sqrt{3} \text{ m}.

From figure,

CD = CE + DE = AB + DE = 7+73=7(1+3)7 + 7\sqrt{3} = 7(1 + \sqrt{3}) m.

Hence, height of tower = 7(1+3)7(1 + \sqrt{3}) meters.

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