Q: If θ_{1} and θ_{2} be the apparent angles of dip observed in two vertical planes at right angles to each other , then the true angle of dip θ is given by :

(a) tan^{2}θ = tan^{2}θ_{1} + tan^{2}θ_{2}

(b) cot^{2}θ = cot^{2}θ_{1} – cot^{2}θ_{2}

(c) tan^{2}θ = tan^{2}θ_{1} – tan^{2}θ_{2}

(d) cot^{2}θ = cot^{2}θ_{1} + cot^{2}θ_{2}

Ans: (d)

Sol: $\large tan\theta_1 = \frac{tan\theta}{cos\alpha} $

$\large cos\alpha = \frac{tan\theta}{tan\theta_1} $ …(i)

$\large tan\theta_2 = \frac{tan\theta}{cos(90-\alpha)} $

$\large tan\theta_2 = \frac{tan\theta}{sin\alpha} $

$\large sin\alpha = \frac{tan\theta}{tan\theta_2} $ …(ii)

$\large sin^2\alpha + cos^2\alpha = 1 $

cot^{2}θ = cot^{2}θ_{1} + cot^{2}θ_{2}