Question

A tower in a city b 150 m high and a multistoreyed hotel at the city center is 20 m high. The angle of elevation of the top of tower from the top of the hotel is 5°. A building, h m high is situated on the straight road connecting the tower with the city center at a distance of 1.2 km from the tower. If the top of the hotel, the top of the building and the top of the tower are in a straight line, then find the height of building h and distance of tower from the city center. (tan 5° = 0.0875, tan 85° = 11.43)

Answer 1

Let AB is a tower of height AB = 150 m. CD is a building of height h m. and EF is a hotel of height EF = 20 m. 

Distance of building from tower BD = 1.2 km = 1200 m.

Let DF = x m

Given, ∠AEP = 5°

From figure PE = BF = (1200 + x)m

From right angled ∆APE

(1200 + x)(0.0875) = 130

⇒ 1200 × 0.0875 + 0.0875x = 130

⇒ 105 + 0.0875x = 130

⇒ 0.0875x = 130

⇒ 0.0875x = 130 – 105 = 25

⇒ x = 25/0.0875

= 286 m (approx)

x = DF = QE

Again from right angled ∆CQE

tan5° = CQ/QE

CQ = QE tan 5°

= 286 × 0.0875

= 25.025 m

Hence, height of building h = CD

= QD + QC

= 20 + 25.025

= 45 m (approx.)

Distance of tower BF = BD + DE

= BD + QE [∵ DF = QE]

= 1200 + 286

= 1486 m

Hence, distance of tower from the city center = 1486 m.