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The angle of elevation of the top of a tower from a point at a distance of 100 metres from its foot on a horizontal plane is found to be 60°. Find the height of the tower

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Let the height of the tower AC be h metres. Given, distance BC = 100 m, ∠ABC = 60°.

∴ tan 60° = \(\frac{AC}{BC} = \frac{h}{100}\) 

⇒ h = 100 x tan 60° = (100 x √3) metre 

= 100 × 1.732 = 173.2 m

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