Question

The magnetic bearing of a line AB was $N{59}^{\circ }{30}^{\prime }W$ in the year 1967, when the declination was $4^{\circ }10E$. If the present declination is $3^{\circ }W$, the whole circle bearing of the line is

• A. ${307}^{\circ }{40}^{\prime }$
• B. ${299}^{\circ }{20}^{\prime }$
• C. ${293}^{\circ }{20}^{\prime }$
• D. ${301}^{\circ }{40}^{\prime }$

Answer 1

The correct option is A ${307}^{\circ }{40}^{\prime }$


$M{N}_1\to \text{old magnetic north}$

$M{N}_2\to \text{present magnetic north}$

Whole circle bearing of line AB from magnetic north

$={360}^{\circ }-{59}^{\circ }{30}^{\prime }$

$={300}^{\circ }{30}^{\prime }$

$Truebearing=Magneticbearing+declination$

$={300}^{\circ }{30}^{\prime }+4^{\circ }{10}^{\prime }$

$={304}^{\circ }{40}^{\prime }$

At present,

$declination=3^{\circ }W=-3^{\circ }$

$WCBofline\left(Magneticbearing\right)=Truebearing-declination$

$={304}^{\circ }{40}^{\prime }-(-3^{\circ })$

$={307}^{\circ }{40}^{\prime }$