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The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second tower is 30° and 24°, respectively. Find the height of the two towers. Give your answer correct to 3 significant figures.

The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second tower is 30° and 24°, respectively. Find the height of the two towers. Give your answer correct to 3 significant figures. Heights and Distances, Concise Mathematics Solutions ICSE Class 10.

Heights & Distances

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Answer

From figure,

AB is the first tower and CD is the second tower.

The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second tower is 30° and 24°, respectively. Find the height of the two towers. Give your answer correct to 3 significant figures. Heights and Distances, Concise Mathematics Solutions ICSE Class 10.

From figure,

EC = BD = 120 m.

In △AEC,

tan 30°=PerpendicularBase13=AEECAE=EC3AE=1203AE=1201.732=69.28 m.\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AE}{EC} \\[1em] \Rightarrow AE = \dfrac{EC}{\sqrt{3}} \\[1em] \Rightarrow AE = \dfrac{120}{\sqrt{3}} \\[1em] \Rightarrow AE = \dfrac{120}{1.732} = 69.28 \text{ m}.

In △EBC,

tan 24°=PerpendicularBase0.445=EBECEB=EC×0.445EB=120×0.445=53.4 m.\text{tan 24°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.445 = \dfrac{EB}{EC} \\[1em] \Rightarrow EB = EC \times 0.445 \\[1em] \Rightarrow EB = 120 \times 0.445 = 53.4 \text{ m}.

From figure,

⇒ CD = EB = 53.4 meters.

⇒ AB = AE + EB = 69.28 + 53.4 = 122.68 ≈ 123 meters.

Hence, height of two towers = 123 meters and 53.4 meters.

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