To solve this problem, we need to find the true angle of dip $\theta$ given the apparent angles of dip $\theta_1$ and $\theta_2$ in two vertical planes at right angles to each other.$\\$The relationship between the true angle of dip and the apparent angles of dip is given by the formula:$\\ \cot^2\theta = \cot^2\theta_1 + \cot^2\theta_2 \\$Let's derive this formula step-by-step:$\\$1. Consider the magnetic field components in the two planes:$\\$• Let $H$ be the horizontal component of the Earth's magnetic field.$\\$• Let $V$ be the vertical component of the Earth's magnetic field.$\\$2. The apparent angles of dip $\theta_1$ and $\theta_2$ are observed in two vertical planes at right angles.$\\$3. The tangent of the apparent angle of dip is given by:$\\ \tan\theta_1 = \frac{V}{H_1} \quad$and$\quad \tan\theta_2 = \frac{V}{H_2} \\$4. The horizontal components $H_1$ and $H_2$ are related to the true horizontal component $H$ by:$\\ H_1 = H \cos\alpha \quad$and$\quad H_2 = H \sin\alpha \\$5. The true angle of dip $\theta$ is given by:$\\ \tan\theta = \frac{V}{H} \\$6. Using the identities for cotangent:$\\ \cot\theta = \frac{H}{V}, \quad \cot\theta_1 = \frac{H_1}{V}, \quad \cot\theta_2 = \frac{H_2}{V} \\$7. Substitute the expressions for $H_1$ and $H_2: \\ \cot\theta_1 = \frac{H \cos\alpha}{V}, \quad \cot\theta_2 = \frac{H \sin\alpha}{V} \\$8. The true angle of dip is related to the apparent angles by:$\\ \cot^2\theta = \left(\frac{H}{V}\right)^2 = \left(\frac{H \cos\alpha}{V}\right)^2 + \left(\frac{H \sin\alpha}{V}\right)^2 \\$9. Simplify the expression:$\\ \cot^2\theta = \cot^2\theta_1 + \cot^2\theta_2 \\$Therefore, the correct option is Option 4:$\\ \cot^2\theta = \cot^2\theta_1 + \cot^2\theta_2 \\$This concludes the derivation and confirms that the true angle of dip is given by the sum of the squares of the cotangents of the apparent angles of dip.