At a certain location in Africa,a compass points $12^{\circ}$ west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points $60^{\circ}$ above the horizontal. The horizontal component of the earth's field is measured to be $0.16 \; G$. Specify the direction and magnitude of the earth's field at the location.

(C) Given:
Angle of declination,$\theta = 12^{\circ}$ West.
Angle of dip,$\delta = 60^{\circ}$ (above the horizontal).
Horizontal component of Earth's magnetic field,$B_{H} = 0.16 \; G$.
We know that the horizontal component $B_{H}$ is related to the total Earth's magnetic field $B$ by the formula:
$B_{H} = B \cos \delta$
Therefore,the magnitude of the Earth's magnetic field $B$ is:
$B = \frac{B_{H}}{\cos \delta} = \frac{0.16}{\cos 60^{\circ}}$
Since $\cos 60^{\circ} = 0.5$,we get:
$B = \frac{0.16}{0.5} = 0.32 \; G$
Direction:
The Earth's magnetic field lies in the vertical plane,$12^{\circ}$ West of the geographic meridian,making an angle of $60^{\circ}$ upwards with the horizontal direction.