The shadow of a tower is found to be x metre longer, when the angle of elevation of the sun changes from 60^ ^ to 45^ ^ . If the height of the tower is 5(3+ 3 ) m , then what is x equal to?

Correct answer: B. 10 m Shadow length when elevation of Sun is 60^ ^ =B C From A B C , B C=A B · (60^ ^ )=5(3+ 3 ) · 1 3 =5( 3 +1) m Shadow length when elevation of Sun is 45^ ^ =B D From A B D , B D=A B · 45^ ^ =5(3+ 3 ) · 1=5(3+ 3 ) m x=B D-B C=5(3+ 3 )-5( 3 +1)=10 m

NDA 2021

Exam: NDA
Year: 2021
Subject: Mathematics
Chapter: Heights and Distances

NDA 2021 · Mathematics · Heights and Distances

The shadow of a tower is found to be x metre longer, when the angle...

The shadow of a tower is found to be

x

metre longer, when the angle of elevation of the sun changes from

60^{{}^{\circ}}

45^{{}^{\circ}}

. If the height of the tower is

5(3+\sqrt{3}) \mathrm{m}

, then what is

x

equal to?

Options

A $8 m$
B $10 m$
C $12 m$
D $15 m$

Correct answer

B. $10 m$

Step-by-step solution

Shadow length when elevation of Sun is

60^{{}^{\circ}}=B C

From

\triangle A B C

B C=A B \cdot \cot \left(60^{{}^{\circ}}\right)=5(3+\sqrt{3}) \cdot \frac{1}{\sqrt{3}}=5(\sqrt{3}+1) \mathrm{m}

Shadow length when elevation of Sun is

45^{{}^{\circ}}=B D

From

\triangle A B D

B D=A B \cdot \cot 45^{{}^{\circ}}=5(3+\sqrt{3}) \cdot 1=5(3+\sqrt{3}) \mathrm{m}\therefore x=B D-B C=5(3+\sqrt{3})-5(\sqrt{3}+1)=10 \mathrm{ m}

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