(a) acute angle outside triangle at point Q = 70°. (Cointerior angles on parallel lines are supplementary (sum up to 180°) Therefore:
Q = 180-110 = 70°
Since angles on a point add up to a revolution (360°), therefore:
200° + 70° + ∠PQR = 360°. i.e, ∠PQR = (360-70-200)° = 90°
(b) bearing from P to R = 110° + ∠QPR. Since we know that ∠PQR = 90°, this is therefore a right angle triangle and we can use trigonometry to solve it.
tan(∠PQR) = 120/150
∠PQR = tan⁻¹(120/150) = 38° 40' = 39°
Hence, the bearing of port R from port P = 110 +39 = 149°T