Question

An observer, 1.5m tall, is 28.5m away from a tower 30m high. Determine the angle of elevation of the top of the tower from his eye. 

Answer 1

Here, ED is the height of the observer and AC is the tower. 
BE = CD = 28. 5 m 
AB = AC - BC = 30 m - 1.5 m = 28.5 m 
In ΔABE, 

`tan <"ABE" = "AB"/"BE"`

⇒ `tan <"ABE" = (28.5"m")/(28.5"m") = 1`

But, `tan 45^circ = 1`

`therefore <"ABE" = 45^circ`

Thus , the required angle of elevation is `45^circ`,