Question

A boy observes the tip of a tower fixed on the top of a building of
height 14m from a point on the ground, then the angle of elevation
is 45º. While walking towards the building again he observes the
tip and base of the tower from another point, now if angles of
elevation are 60°and 30° respectively. Find the height of the tower
and the distance he walked.​

Answer 1

Step-by-step explanation:

elevation

is 45º. While walking towards the building again he observes the

tip and base of the tower from another point, now if angles of

Answer 2

Height of the tower is 56m(42+14) and the distance boy walked is 17.752m

• When the angles are 60° and 30° the height of the boy from the ground is 14m so the horizontal distance of the boy from tower is given by tan(30°)=14/b (i.e. b=24.2487m) (see the ΔABD)
• In ΔBCD tan(60°)=a/24.2467 ⇒a= 42m
• In ΔOBC OC=OB as ∠COB=45° ⇒OD=42-24.2487=17.75m(distance the boy moved)

Please refer to the picture provided below