math question urgent

[email protected] SG

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

^tamago^ SG

Oh no not this relative velocity question again. Neutral Laughing

[email protected] SG

not velocity question la is trigo la

teraexa SG

Bah I shall attempt this question.

Assuming bearing from P = clockwise angle from due north.

All distances are in km and angles in degrees.

Let's have this point named X that is due north of P (bearing 0) and collinear with Q and R (ie X, Q and R) are in one line.

Distance XQ = 1.2 sin 30 = 0.6 km
Distance PX = sqrt(1.2^2 - 0.6^2) = 1.2 cos 30 = 1.039 km
Distance QR = 12/60 x 5 = 1km
Distance of PR = sqrt(PX^2 + XR^2) = sqrt(1.039^2 + 1^2) = 1.442 km

Let bearing of R from P be alpha.
tan alpha = XR/PX = 1.6/1.039
alpha = 57.0 deg, therefore bearing of R from P = 057

Vertical height of balloon = 1.2 tan 21 = 0.4606 km
Let angle of depression of P from balloon over R be beta.
Using alternate angles, angle of depression of P from balloon = angle of elevation of balloon from P.
tan beta = 0.4606/PR = 0.4606/1.442
beta = 17.7 deg

Double check my working. Been out of practice for a fortnight already.

gunner77 SG

wah sian

i hate maths Rolling Eyes