Question

From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer 1

Let AB be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°
EC = DE + CD
also, CD = AB = 7 m.
and BC = AD
A/q,
In  right ΔABC,
tan 45° = AB/BC
⇒ 1= 7/BC
⇒ BC = 7 m = AD
also,
In  right ΔADE,
tan 60° = DE/AD
⇒ √3 = DE/7
⇒ DE = 7√3 m
Height of the tower = EC =  DE + CD
                                = (7√3 + 7) m = 7(√3+1) m.