A radio antenna stands on top of a building 300 m tall. At a point on the ground, the angle of elevation of bottom of antenna is 45° and the top of antenna is 60°. What is the height of the antenna? -

219.6 m 

height of building = BC m , height of antenna = AB m

In triangle BCD

tan 45° = $\frac{BC}{CD}$

1 = $\frac{BC}{CD}$

As , BC = 300 m . so , BD is also 300 m

in triangle ACD

tan 60° =  $\frac{AC}{CD}$

$\sqrt{3}$  = $\frac{AC}{CD}$

$\sqrt{3}$  = AC , CD = 300

So  , AC =  300$\sqrt{3}$

AB = AC - BC

     = 300$\sqrt{3}$ - 300 = 300(1.732 - 1) = 300 x 0.732 = 219.6 m