The angles of elevation of the top of a tower from two places situated at distances 21 m and x m from the base of the tower are 45^ ^ and 60^ ^ respectively. What is the value of x ?

Correct answer: A. 7 3 Let A B is a tower, Given that AC=21 m and AD=x m From ABC, (45^ ^ )= AB AC 1= AB 21 A B=21 m Again, from ABD, aligned (60^ ^ ) & = AB AD 3 =(21)/(x) x & =7 3 m aligned

NDA 2013

Exam: NDA
Year: 2013
Subject: Mathematics
Chapter: Heights and Distances

NDA 2013 · Mathematics · Heights and Distances

The angles of elevation of the top of a tower from two places...

The angles of elevation of the top of a tower from two places situated at distances 21 m and

x \mathrm{ m}

from the base of the tower are

45^{{}^{\circ}}

and

60^{{}^{\circ}}

respectively. What is the value of

x

Options

A $7 \sqrt{3}$
B $7-\sqrt{3}$
C $7+\sqrt{3}$
D $14$

Correct answer

A. $7 \sqrt{3}$

Step-by-step solution

Let

A B

is a tower, Given that

\mathrm{AC}=21 \mathrm{ m}

and

\mathrm{AD}=x \mathrm{ m}

From

\triangle \mathrm{ABC}, \tan \left(45^{{}^{\circ}}\right)=\frac{\mathrm{AB}}{\mathrm{AC}}1=\frac{\mathrm{AB}}{21}\therefore A B=21 \mathrm{ m}

Again, from

\triangle \mathrm{ABD},

\begin{aligned} \tan \left(60^{{}^{\circ}}\right) & =\frac{\mathrm{AB}}{\mathrm{AD}} \Rightarrow \sqrt{3}=\frac{21}{x} \\ x & =7 \sqrt{3} \mathrm{ m} \end{aligned}

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