When positioned 10 cm from the needle, a short magnet causes a 37^...

In the Tan-A position, the magnetic field at the needle due to a short magnet at distance $d_1$ is given by:

$B_A = \dfrac{\mu_0}{4\pi} \dfrac{2M}{d_1^3}$

The deflection $\theta$ is given by $B_A = B_H \tan \theta$.

In the Tan-B position, the magnetic field at the needle due to the same short magnet at distance $d_2$ is given by:

$B_B = \dfrac{\mu_0}{4\pi} \dfrac{M}{d_2^3}$

The deflection $\theta$ is given by $B_B = B_H \tan \theta$.

Since the deflection is 37^{}^{{}^{\circ}} in both cases, the magnetic fields must be equal:

$B_A = B_B$

$\dfrac{\mu_0}{4\pi} \dfrac{2M}{d_1^3} = \dfrac{\mu_0}{4\pi} \dfrac{M}{d_2^3}$

$\dfrac{2}{d_1^3} = \dfrac{1}{d_2^3}$

$d_2^3 = \dfrac{d_1^3}{2}$

$d_2 = \dfrac{d_1}{2^{1/3}}$

Given $d_1 = 10 \text{ cm}$ and using $2^{1/3} \approx 1.26$:

$d_2 = \dfrac{10}{1.26} \approx 7.9 \text{ cm}$