A boy is looking at the top of a building which is 10 m high.
From the figure, the distance between the boy and the building will be .

10 m

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20 m

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5 m

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25 m

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Solution

The correct option is A 10 m

From the above figure, we can write,
$∠ X B A = ∠ B A C = 45 °$
$[Alternate interior angles]$

$Therefore, i n △ A C B,$

$t a n A = \frac{B C}{A C}$
$\Rightarrow t a n 45 ° = \frac{10}{A C}$
$\Rightarrow 1 = \frac{10}{A C}$
$\Rightarrow A C = 10 m$

Hence, the distance between the boy and the building will be 10 m.

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A boy is looking at the top of a building which is 10 m high. From the figure, the distance between the boy and the building will be .

The correct option is A 10 m From the above figure, we can write, ∠XBA=∠BAC=45° [Alternate interior angles] Therefore, in △ ACB, tanA=BCAC ⇒tan 45°=10AC ⇒1=10AC ⇒AC=10 m Hence, the distance between the boy and the building will be 10 m.

A boy is looking at the top of a building which is 10 m high.
From the figure, the distance between the boy and the building will be .