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If the whole circle bearing of a line is 150°, its reduced bearing will be: ______.
1. S
2. W
3. S 60° E
4. S 30° E

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Correct Answer - Option 4 : S 30° E

Concept:

The relation between WCB and RB is given in the table below:

S.no.

Whole circle bearing (WCB)

Reduced bearing (RB)

Quadrant

1

\(0^\circ - 90^\circ\)

WCB

NE

2

\(90^\circ - 180^\circ\)

180°- WCB 

SE

3

\(180^\circ - 270^\circ\)

WCB - 180° 

SW

4

\(270^\circ - 360^\circ\)

\({\rm{36}}{{\rm{0}}^{\rm{^\circ }}}{\rm{ - WCB}}\)

NW

Calculation:

Given,

WCB of a line = 150° 

If the value of WCB lies between 90° to 180°, then,

Reduced bearing  = 180° - WCB and the line will be in South direction

∴ RB = 180° - 150° = 30° 

Hence the reduced bearing of a line is S 30° E.

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