The bearing of Q from P is150 and the bearing of P from R is 015. If Q and R are 24km and 32km respectively from P. Represent this information into a diagram. Calculate the distance between Qand R. Find the bearing of R from Q.

Answer 1

I hope you made your sketch.
by vectors
vector RQ = (32cos15°, 32sin15°) + (24cos150, 24sin150°)
= (30.9096..., 8.2822...) + (-20.7846.., 12)
= (10.125.. , 20.2822...)

maginitude = √(10.125..^2 + 20.2822...^2) = appr 22.669

or
using the cosine law:
my diagram had a triangle of sides 32 and 24 with an angle of 45 ° between them
PR^2 = 24^2 + 32^2 - 2(24)(32)cos45°
= 576 + 1024 - 1086.116..
= 513.8839...
PR = √513.8839... = appr 22.669 , just like above

Answer 2

All angles are measured CW from +y-axis.
Given: PQ = 24km[150o], RP = 32km[15o].
a. RQ = RP + PQ = 32[15o] + 24[150o]
RQ = (32*sin15+24*sin150) + (32*cos15+24*cos150)I
RQ = 20.3 + 10.1i. = 22.7km[63.5o].

b. QR = 22.7[63.5+180] = 22.7km[243.5o].