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 $x$ ?

• Correct answer
  A

  $7 \sqrt{3}$
• B

  $7-\sqrt{3}$
• C

  $7+\sqrt{3}$
• D

  $14$

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

Hence, the correct answer is option: 'A'.