Question

18. Rishu is riding in a hot air balloon. After reaching a point P, he spots a car parked at B on
the ground at an angle of depression of 30°. The balloon rises further by 50 metres und
now he spots the same car at an angle of depression of 45° and a lorry parked at B’ at an
angle of depression of 30°. (Use 3 = 1.73)
P
个个个
Rishu
Р
45°
Lorry
30°
"Car
B'
B​

Answer 1

Answer:

The angle of elevation of a hot air balloon, climbing vertically, changes from 25 degrees at 10:00 am to 60 degrees at 10:02 am. The point of observation of the angle of elevation is situated 300 meters away from the take off point. What is the upward speed, assumed constant, of the balloon? Give the answer in meters per second and round to two decimal places.

Answer 2

Given :- Rishu is riding in a hot air balloon. After reaching a point P, he spots a car parked at B on the ground at an angle of depression of 30°. The balloon rises further by 50 metres and now he spots the same car at an angle of depression of 45° .

To Find :- How far is the car from the base of of ballon ?

Solution :-

from image let,

• P = Initial position of Ballon .
• B = car .
• PQ = 50 m .

so,

In right ∆PAB we have,

→ tan 30° = PA / AB

→ (1/√3) = PA/AB

→ AB = √3•PA --------- Eqn.(1)

now, In right ∆QAB we have,

→ tan 45° = QA / AB

→ 1 = QA/AB

→ AB = QA

→ AB = (QP + PA)

→ AB = (50 + PA)

putting value of Eqn.(1) ,

→ √3PA = 50 + PA

→ √3PA - PA = 50

→ PA(√3 - 1) = 50

→ PA(1.73 - 1) = 50

→ PA = 50/(0.73)

therefore,

→ AB = √3•PA

→ AB = (1.73 * 50) / 0.73

→ AB ≈ 118.5 m (Ans.)

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