Question

A $1.5$ metre tall boy saw the top of a building under construction at an elevation of ${30}^o$. The completed building was $10$ metres higher and the boy saw its top at an elevation of ${60}^o$ from the same spot. What is the height of the building?

Answer 1

Based on the given information, we can drawn the figure shown above.

Here, $AB=1.5$

$CD=10$

Let $AB=x$

$CD=y$

$\therefore \tan {30}^o=\frac{CO}{AO}-AB$

$\therefore \frac{1}{\sqrt{3}}=\frac{y}{x}-1.5$

$\therefore x=\sqrt{3}(y-1.5)$

$\therefore \tan {60}^o=\frac{DO-AB}{AO}$

$\therefore \sqrt{3}=\frac{y+10-1.5}{x}$

$\therefore \sqrt{3}x=y+8.5$

$\therefore \sqrt{3}\left(\sqrt{3}(y-1.5)\right)=y+8.5$

$\therefore 3y-4.5=y+8.5$

$\therefore 2y=13$

$\therefore y=6.5$

$\therefore$ Height of building $=6.5+10=16.5m$