Question

The whole circle bearing of a line is ${287}^0{15}^{\prime }$, The reduced bearing of the line is :

• A. $\text{S}{107}^0{15}^{\prime }\text{W}$
• B. $\text{S}{17}^0{15}^{\prime }\text{W}$
• C. $\text{S}{107}^0{15}^{\prime }\text{E}$
• D. $\text{N}{72}^0{45}^{\prime }\text{W}$

Answer 1

The correct option is D $\text{N}{72}^0{45}^{\prime }\text{W}$

(c)
Conversion of whole circle bearing of a line $={287}^0{15}^{\prime }$

R.B $={360}^0-{287}^0{15}^{\prime }={72}^0{15}^{\prime }$

Quadrant = NW

So, ans is ${72}^0{45}^{\prime }$ W.