At a location where the angle of dip is 60^ ^ ^ , the earth's...

Given angle of dip, \delta = 60^{}^{{}^{\circ}}

Horizontal component of earth's magnetic field, $B_H = 26 \mu\text{T}$

The vertical component $B_V$ is given by:

$B_V = B_H \tan\delta$

B_V = 26 \tan 60^{}^{{}^{\circ}} = 26 \sqrt{3} \approx 26 \times 1.732 = 45.03 \mu\text{T} \approx 45 \mu\text{T}

The magnitude of the earth's magnetic field $B$ is given by:

$B = \dfrac{B_H}{\cos\delta}$

B = \dfrac{26}{\cos 60^{}^{{}^{\circ}}} = \dfrac{26}{1/2} = 52 \mu\text{T}

Thus, the vertical component is $45 \mu\text{T}$ and the magnitude of the magnetic field is $52 \mu\text{T}$.