Question

At a certain point the angle of elevation of a tower is found to be ${\cot }^{-1}\frac{3}{5}$. On walking $32m$ directly towards the tower its angle of elevation is ${\cot }^{-1}\frac{2}{5}$ . The height of the tower (in metres) is

• A. $32$
• B. $160$
• C. $320$
• D. $340$

Answer 1

Answer: (B)

$160$

$\cot {\theta }_1=\frac{3}{5}$

$\cot {\theta }_2=\frac{2}{5}$

$\cot {\theta }_1=\frac{(32+a)}{h}$

$\cot {\theta }_2=\frac{a}{h}$

$\Rightarrow a=\frac{2}{5}h$

$\frac{3}{5}=\frac{32+\frac{2}{5}h}{h}$

$\Rightarrow \frac{1}{5}=\frac{32}{h}$

$\Rightarrow h=160m.$