Section 4: Soil and Water Conservation Engineering
Surveying and Levelling
Subtopic: Measurement of angles and bearings
Question Statement:
The Whole Circle Bearing (WCB) of a traverse line is $245^\circ 30'$. The numeric angle value of its Reduced Bearing (quadrantal bearing) in degrees is ________ (round off to 1 decimal place).
Answer Key:
65.5
Step-by-Step Solution & Explanation
Since the bearing lies between $180^\circ$ and $270^\circ$ (third quadrant, South-West):
$\text{Reduced Bearing} = \text{WCB} - 180^\circ$
$\text{RB} = 245^\circ 30' - 180^\circ 00' = 65^\circ 30'$
In decimal degrees:
$65^\circ + \frac{30'}{60} = 65.5^\circ$
(The quadrant designation is $\text{S } 65^\circ 30' \text{ W}$).