Question

Two men are on opposite side of 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 meters, find the distance between the two men.

Answer 1

Let CD be the tower and A and B be the positions of the two men standing on the opposite sides.
Thus, we have:
∠DAC = 30°,  ∠DBC = 45° and CD = 50 m
Let AB =  xmand BC =  ymsuch that AC = (x - y)m.

In the right ΔDBC,we have:

${\displaystyle \frac{CD}{BC}=\tan 45°=1}$

${\displaystyle \Rightarrow \frac{50}{y}=1}$

⇒  y = 50 m

In the right  ΔACD,we have:

${\displaystyle \frac{CD}{AC}=\tan 30°=\frac{1}{\sqrt{3}}}$

${\displaystyle \Rightarrow \frac{50}{(x-y)}=\frac{1}{\sqrt{3}}}$

${\displaystyle x-y=50\sqrt{3}}$

On putting y  = 50in the above equation, we get:

${\displaystyle x-50=50\sqrt{3}}$

${\displaystyle \Rightarrow x=50+50\sqrt{3}=50(\sqrt{3}+1)=136.6m}$

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