Question

The angle of elevation of the top of an unfinished tower at a point 150 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60°? 

Answer 1

Let BC be the length of unfinished tower. Let the tower be raised upto point A so that the angle of elevation at point A is 60°. D is the point on ground from where elevation angles are measured. 

In ΔBCD

`"BC"/"CD" = tan 30^circ`

`"BC"/"CD" = 1/sqrt(3)`

`"BC" = "CD"/sqrt(3)`

`"BC" = 150/sqrt(3)`   ... (1)

In ΔACD

`("AB + BC")/("CD") = tan 60^circ`

⇒ `("AB + BC")/("CD") = sqrt(3)`

⇒ `("AB" + 150/sqrt(3))/150 = sqrt(3)`  ....(Using (i))

⇒ `"AB" = 150sqrt(3) - 150/sqrt(3) = (150 xx 3 - 150)/sqrt(3)`

⇒ `"AB" = 300/sqrt(3) = 300/1.732 = 173.2` m

Thus , the required height is 300 m.