A magnetic needle free to rotate in a vertical plane

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its North tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 gauss. Determine the magnitude of the earth’s magnetic field at the place.

Angle of dip, δ = 60° = $\pi$/3
Horizontal component of the earth’s magnetic field, H = 0.4 gauss
Earth magnetic field, $B_{ e }$ = ?
We know that, H = $B_{ e }$ cos δ
.•. Horizontal component of the earth’s magnetic field,
=> $B_{ e }$ = H/cos δ = 0.4/cos 60°
= 0.4 /1/2 = 0.8
$B_{ e }$ = 0.8 gauss