The bearings of P and Q from a common point N are 020° and 300° respectively. If P and Q are also equidistant from N, find the bearing of P from Q.
- 040°
-
070°
- 280°
- 320°
The correct answer is: B
Explanation
From the diagram above, <PNQ = 80º
But < PQN = <NPQ = 50º ( base <s of isosceles triangle PQN)
Therefore, the bearing of P from Q = 070º
There is an explanation video available .