Question

From the top of a $7$m high building, the angle of elevation of the top of a cable tower is ${60}^o$ and the angle of depression of its foot is ${45}^o$. The height of the tower in metre is

• A. $7(\sqrt{3}-1)$
• B. $7\sqrt{3}$
• C. $7+\sqrt{3}$
• D. $7(\sqrt{3}+1)$

Answer 1

The correct option is D $7(\sqrt{3}+1)$

In $\Delta$ABE,
$\tan {60}^o=\frac{h}{AB}$
$h=AB\sqrt{3}$ ....(i)
In $\Delta$ABC,
$\tan {45}^o=\frac{7}{AB}$
$AB=7$ ...(ii)
By equation (i) and equation (ii)
$h=7\sqrt{3}$
So, height of tower is $=h+7=7(\sqrt{3}+1)$