Pqr is allocated in the same horizontal plain, the bearing of q from p is130° and the bearing of r from q is 075°, then pq=6cm and qr =4cm, find the bearing of r from p, collect to the nearest degree

Answer 1

to use law of cosines we need angle PQR
PQ is 40deg down from x direction (East)
QR is 15 deg up from x direction (East)
so angle PQR is 180 - 40 - 14 = 126 deg
then
q^2 = p^2 + r^2 - 2 p r cos 126
q^2 = 4^2 + 6^2 - 2*4*6 cos 126
solve for q
then get the other angles from law of sines
eg:
sinQPR / 4 = sin 126 / q

Answer 2

All angles are measured CW from +y-axis,
PR = PQ + QR = 6cm[130o] + 4cm[75o],
X = 6*sin130 + 4*sin75 = 8.46 cm,
Y = 6*Cos130 + 4*Cos75 = -2.82 cm,

TanA = X/Y,
A = -71.6o = 71.6o E. of S. = 108o,
The bearing of R from P = 108o.