At a place, a dip circle and a magnetometer show a horizontal part of 28 µT and a vertical part of…

Step 1. The whole field. $B_E = \sqrt{(28\ \mu\text{T})^2 + (33\ \mu\text{T})^2} = \sqrt{1873}\ \mu\text{T} = 43.3\ \mu\text{T}$.

Step 2. The dip. $\tan\delta = \dfrac{33\ \mu\text{T}}{28\ \mu\text{T}} = 1.179$. So $\delta = 49.7^\circ$.

Step 3. The horizontal part at the second place. $B_H = 47\ \mu\text{T} \times \cos 62^\circ = 22.1\ \mu\text{T}$.

Step 4. The vertical part at the second place. $B_V = 47\ \mu\text{T} \times \sin 62^\circ = 41.5\ \mu\text{T}$.