Question

If whole circle bearing of a line is ${210}^0{0}^{\prime }{0}^{\prime \prime },$ its value in quadrantal bearing system is:

• A. S ${30}^0$ 0' 0 " E
• B. N ${30}^0$ 0' 0 " W
• C. N ${30}^0$ 0' 0 " E
• D. S ${30}^0$ 0' 0 " W

Answer 1

The correct option is D S ${30}^0$ 0' 0 " W

(a)
WCB of a line = ${210}^0{0}^{\prime }{0}^{\prime \prime }$
Reduced bearing = ${210}^0{0}^{\prime }{0}^{\prime \prime }$ - ${180}^0{0}^{\prime }0"$ = ${30}^0{0}^{\prime }{0}^{\prime \prime }$
Quadrant = SW
Bearing in Q.B system = S ${30}^0$ 0' 0" W