Question

A man observes the elevation of a tower to be ${30}^{\circ }$. After advancing $11\text{cm}$ towards it, he finds that the elevation is ${45}^{\circ }$. The height of the tower to the nearest meter is

• A. $10\text{m}$
• B. $15\text{m}$
• C. $20\text{m}$
• D. $22\text{m}$

Answer 1

The correct option is B $15\text{m}$

Let $OT$ be the tower of height $h$

Given:

$PQ=\text{11 m}$

Since $\angle OQT={45}^{\circ }$

From, $△OQT,$

$\tan {45}^{\circ }=\frac{OT}{OQ}$

$⟹1=\frac{OT}{OQ}$

$⟹OT=OQ$
$\therefore OT=OQ=h$
From $△OPT,\frac{OP}{OT}=\cot {30}^{\circ }$
or $\frac{h+11}{h}=\sqrt{3}$
or $h=\frac{11}{\sqrt{3}-1}=\frac{11}{2}(\sqrt{3}+1)$ (on rationalisation)
or $h=1.366\times 11$
$=15.026\approx 15$