A tower's angle of elevation from a point is $30°$. The angle of elevation is $60°$at a point on the horizontal line passing through the foot of the tower and 50 metres closer to it. The distance is

The angle of elevation is the angle between the horizontal line of sight and the object when a person stands and looks up at something.

The angle of depression is the angle between the horizontal line of sight and the object if a person stands and looks down at an object.

Let$Ac=xm$ and $AB=hm$from $∆\mathrm{BCA}$

$\tan 30°=\frac{AB}{AC}$ or $\frac{1}{\sqrt{3}}=\frac{h}{x}$ or $h=\frac{x}{\sqrt{3}}$

From $∆\mathrm{BDA} \tan 60°=\frac{\mathrm{AB}}{\mathrm{AD}}$

Or $\sqrt{3}=\frac{H}{x-50}$ or $h=(x-50)\sqrt{3}$

Equation the two values of h, we get

$\frac{x}{\sqrt{3}}=(x-50)\sqrt{3}$

or $2\mathrm{x}=150$ therefore$\mathrm{x}=75 \mathrm{m}$