P, Q and R are 3 points on the same horizontal ground. The bearing of Q from P is 030° and PQ is 1200 m apart and hot air balloon is spotted vertically above Q at an angle of elevation of 21° from P. The balloon is drifting at a constant height towards the east at a steady speed of 12 km/h. Five minutes after passing Q , it is observed to be directly above R . Find
1) The distance between P and R
2) The bearing of R from p
3) the angle of depression of P from the balloon when it is directly above R
Answer:
1) After 5 minutes, the balloon would have travelled east at 12 km/h× 5/60 h = 1 km
So if we draw out the points P, Q and R, it will look like the following:
So angle PQR = 90° + 30° = 120°
Using cosine rule,
PR² = PQ² + QR² - 2 × (PQ)(QR)(cos 120°)
PR² = 1.2² + 1² - 2 × (1.2)(1)(-½)
PR² = 3.64
PR = 1.908 km
2) Using sine rule,
(sin ∡QPR ) / 1 = (sin ∡PQR) / 1.908
sin ∡QPR = sin 120 / 1.908
sin ∡QPR = 0.45389
∡QPR = 27°
Hence, bearings of R from P = 030 + 027 = 057
3) The diagrams can be redrawn to have a side view so as to see it clearer
Vertical height of balloon = 1.2 tan 21° = 0.4606 km
Because it is at a constant height, the following diagram is what happened when it reaches R:
Thus,
tan θ = 0.4606 / PR
tan θ = 0.4606 / 1.908
θ = 13.6°
Angle of depression of P from the balloon when it is directly above R = 13.6°