A magnet hung at $$45°$$ with magnetic meridian makes an angle of $$60°$$ with the horizontal. The actual value of the angle of dip is

We need to find the actual angle of dip ($$\delta$$) given the apparent angle of dip ($$\delta'$$) in a plane inclined to the magnetic meridian.

1. Identify the Given Values

• Angle of inclination with the magnetic meridian ($$\alpha$$) = $$45^\circ$$
• Apparent angle of dip ($$\delta'$$) = $$60^\circ$$

2. Formula for Apparent Dip

The relationship between the actual angle of dip ($$\delta$$), the apparent angle of dip ($$\delta'$$), and the angle with the magnetic meridian ($$\alpha$$) is given by the formula:

$$\tan \delta' = \frac{\tan \delta}{\cos \alpha}$$

Rearranging the formula to solve for the actual dip ($$\tan \delta$$):

$$\tan \delta = \tan \delta' \cdot \cos \alpha$$

3. Calculation

Substitute the known trigonometric values ($$\tan 60^\circ = \sqrt{3}$$ and $$\cos 45^\circ = \frac{1}{\sqrt{2}}$$) into the equation:

$$\tan \delta = \sqrt{3} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}}$$

Taking the inverse tangent to isolate $$\delta$$:

$$\delta = \tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$$

Conclusion

The actual value of the angle of dip is $$\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$$.