Question

Two men are on opposite sides of a tower. they measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 metres, find the distance between the two men. $\left[\mathrm{Take}\sqrt{3}=1.732\right]$

Answer 1

Let $CD$ be the tower and $A\mathrm{and}B$ be the positions of the two men standing on the opposite sides. Thus, we have:
∠$DAC={30}^o$, ∠$DBC={45}^o$ and $CD\hspace{0.17em}=50$ m
Let $AB=x$ m and $BC\hspace{0.17em}=y$ m such that $AC=(x-y)$ m.

In the right ∆$DBC$, we have:
$\frac{CD}{BC}=\tan {45}^o=1$

⇒ $\frac{50}{y}=1$
⇒ $y=50$ m

In the right ∆$ACD$, we have:
$\frac{CD}{AC}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{50}{(x-y)}=\frac{1}{\sqrt{3}}$
⇒ $x-y=50\sqrt{3}$
On putting $y=50$ in the above equation, we get:
$x-50=50\sqrt{3}$
⇒ $x=50+50\sqrt{3}=50(\sqrt{3}+1)=136.6$ m

∴ Distance between the two men = $AB=x=136.6$ m