The angle of elevation of the top of a vertical

Solution:

Let the height the vertical tower, OT = H

and $\quad O P=A B=x \mathrm{~m}$

Given that, $A P=10 \mathrm{~m}$

and $\quad \angle T P O=60^{\circ}, \angle T A B=45^{\circ}$

Now, in $\triangle T P O$, $\tan 60^{\circ}=\frac{O T}{O P}=\frac{H}{r}$

$\Rightarrow$ $\sqrt{3}=\frac{H}{x}$

$\Rightarrow$ $x=\frac{H}{\sqrt{3}}$ ...(i)

and in $\triangle T A B$, $\tan 45^{\circ}=\frac{T B}{A B}=\frac{H-10}{x}$

$\Rightarrow$ $1=\frac{H-10}{x} \Rightarrow x=H-10$

$\Rightarrow$ $\frac{H}{\sqrt{3}}=H-10$ [from Eq.(1)]

$\Rightarrow$ $H-\frac{H}{\sqrt{3}}=10 \Rightarrow H\left(1-\frac{1}{\sqrt{3}}\right)=10$

$\Rightarrow$ $H\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)=10$

$\therefore$ $H=\frac{10 \sqrt{3}}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}$ [by rationalisation]

$=\frac{10 \sqrt{3}(\sqrt{3}+1)}{3-1}=\frac{10 \sqrt{3}(\sqrt{3}+1)}{2}$

$\Rightarrow$ $=5 \sqrt{3}(\sqrt{3}+1)=5(\sqrt{3}+3) \mathrm{m} .$

Hence, the required height of the tower is 5 (√3 + 3) m,