c
Given, angle of declination, \(\theta=12^{\circ}\) West Angle of dip \(\delta=60^{\circ}\)

Horizontal component of earth's magnetic field

\(\mathrm{H}=0.16\, \mathrm{G}\)

Let the magnitude of earth's magnetic field at that place is \(\mathrm{R}\)

Using the formula, \(H=R \cos \delta\)

\(\mathrm{R} =\frac{-\mathrm{H}}{\cos \delta}=\frac{0.16}{\cos 60^{\circ}}=\frac{0.16 \times 2}{1}\)

\(=0.32\,\mathrm{G}=0.32 \times 10^{-4}\, \mathrm{T}\)

The earth magnetic field lies in a vertical plane \(12^{\circ}\) West of geographical meridian at an angle \(60^{\circ}\) above the horizontal.