Question

from the top of a building 60m high ,the angle of depressions of top and bottom of a tower are 30°and60°.find the height of the tower ​

Answer 1

ANSWER :

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$\boxed{40\:\: m }$

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EXPLANATION :

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Given :

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Height of building (BC) = 60m

Angle of depression of top (∠ABE) = 30°

Angle of depression of bottom (∠ABD) = 60°

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Solution :

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Refer to attachment for diagram.

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We know alternate angles are equal.

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∠ABE = ∠BEF = 30°

∠ABD = ∠BDC = 60°

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Now,

In △BDC

$\tan(60°) = \frac{BC}{DC}$

☆$\sqrt{3} = \frac{60}{DC}$

☆$DC = \frac{60}{ \sqrt{3} }$

☆$DC = 20 \sqrt{3} \: \: m$

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Since, DCFE is a rectangle so, DC = EF = 20√3m

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In △BEF

$\tan(30°) = \frac{BF}{EF}$

☆$\frac{1}{ \sqrt{3} } = \frac{BF}{20 \sqrt{3} }$

☆$BF = \frac{20 \sqrt{3} }{ \sqrt{3} }$

☆$BF = 20 \: \: m$

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Since, DCFE is a rectangle

ED = FC

☆ED = BC - BF

☆ED = (60 - 20) m

☆ED = 40 m

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Height of tower (ED) = 40 m