Question

The bearing of the line $AB$ from North is $143^\circ 40'$ and angle $ABC$ measured in clockwise direction is $309^\circ 30'$. The bearing of line $BC$ in Quadrantal Bearing System is ____________

• A. N$3^\circ 10'$W
• B. N$86^\circ 50'$E
• C. N$86^\circ 50'$W
• D. N$3^\circ 10'$E

Hint:

At a vertex $B$, first find the {back bearing} of the incoming line. Add the included angle (respecting clockwise/anticlockwise sense) to get the outgoing line’s WCB, then convert to quadrantal form by quadrant rules.

Answer 1

The Correct Option is C

Step 1: Convert the given data to whole-circle bearings (WCB).
WCB of $AB$ is given: $\,\theta_{AB}=143^\circ40'$.
At station $B$, the back bearing of $AB$ equals \[ \theta_{BA}=\theta_{AB}+180^\circ=143^\circ40'+180^\circ=323^\circ40'. \] Step 2: Turn the specified clockwise angle at $B$ to get the WCB of $BC$.
Angle $ABC$ (from $BA$ to $BC$ measured clockwise) is $309^\circ30'$. Hence \[ \theta_{BC} =\theta_{BA}+309^\circ30' =323^\circ40'+309^\circ30' =633^\circ10' \equiv 273^\circ10' \ (\text{mod }360^\circ). \] Step 3: Convert WCB to Quadrantal Bearing (QB).
Since $273^\circ10'$ lies in the NW quadrant ($270^\circ$–$360^\circ$), \[ \text{QB} = \text{N}\big(360^\circ-\theta_{BC}\big)\text{W} = \text{N}\big(360^\circ-273^\circ10'\big)\text{W} = \text{N}86^\circ50'\text{W}. \] \[ \boxed{\text{N}86^\circ50'\text{W}} \]