From the top of a tower 300 {\m} high, the angles of depression of the top of the building and its bottom are observed to be 45^\circ and 60^\circ respectively. Find the height of the building. (Take \sqrt{3}=1.73)

Management Aptitude Test Dec 2019 Paper

Question : 98 of 200

Marks: +1, -0

From the top of a tower

300 {\m}

high, the angles of depression of the top of the building and its bottom are observed to be

45°

and

60°

respectively. Find the height of the building. (Take

√{3}=1.73)

Solution:

Let

\BC

be the tower and

\AD

be the building.
Draw

\DE ⊥\BC

.
Then,

\BC=300 {\m}, ∠\EDC=45°

and

∠ \BAC=60° .

Let

\CE=x

metres.

From right

\Δ\EDC, \tan 45°={\CE}/{\DE}

1=\CE/\DE

⇒\CE=\DE=\AB=x

From right

\Δ A B C, \tan 60°={B C}/{A B}={300}/{x}

√{3}={300}/{x}

⇒ x={300}/{√{3}} {\m}

∴ \BE=\BC-\BE=(300-{300}/{√{3}}) {\m}

=(300-100 √{3}) {\m}

=300-100 ×1.73

=300-173=127 {\m}

But

\AD=\BE

⇒ \AD=127 {\m}

Hence, the height of the building is

127 {\m}

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