A magnetic needle is free to rotate in a vertical plance which makes an angle of `60^@` with the magnetic meridian. If the needle stays in a direction making an angle of `tan^(-1) (2sqrt(3))` with the horizontal, what would be the dip at that place?

Explanation

To solve the problem, we need to find the dip angle (δ) at the location where a magnetic needle is free to rotate. The needle is inclined at an angle of \(60^\circ\) with the magnetic meridian and makes an angle of \(\tan^{-1}(2\sqrt{3})\) with the horizontal.

### Step-by-Step Solution:

1. **Understanding the Angles**:
- The magnetic needle is inclined at an angle of \(60^\circ\) with respect to the magnetic meridian.
- The angle with the horizontal is given as \(\tan^{-1}(2\sqrt{3})\).

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