Question

In the quadrantal bearing system, a whole circle bearing of ${293}^0{30}^{\prime }$ can be expressed as:

Answer 1

(c)
Bearing in WCB system = ${293}^0{30}^{\prime }$
R.B = ${360}^0-{293}^0{30}^{\prime }={66}^0{30}^{\prime }$
Quadrant = NW
Hence bearing in Q.B system = N ${66}^0{30}^{\prime }$ W