Question

The angle of elevation of a tower from a point is ${30}^0.$ At a point on the horizontal line passing through the foot of 'the tower and 50 metros nearer it, the angle of elevation is ${60}^0.$ The distance of the first point from the tower is

• A. 50 metres
• B. 75 metres
• C. 100 metres
• D. 150 metres

Answer 1

Answer: (B)

75 metres

In the above figure let AB be the height of the tower
Let AC be the distance of the first point from the foot of the tower
It is given that $\angle BDA={60}^0$ and $\angle ACB={30}^0$
Now consider triangle BCD
$\angle BDC={180}^0-{60}^0={120}^0$ ...(linear pair)
$\angle CBD={180}^0-({120}^0+{30}^0)={30}^0$
Hence BCD is an isosceles triangle with CD=BD=50m ...(i)
Therefore in triangle ABD
$AD=BDcos{60}^0$
$=\frac{50\left(1\right)}{\left(2\right)}=25m$ ...from(i)
Hence $AC=CD+AD=25m+50m=75m$
Hence answer is B