Question

If ${\theta }_1$ and ${\theta }_2$ be the angles of dip observed in two vertical planes at right angles to each other and $\theta$ be the true angle of dip, then

• A. ${\cos }^2\theta ={\cos }^2{\theta }_1+{\cos }^2{\theta }_2$
• B. ${\sec }^2\theta ={\sec }^2{\theta }_1+{\sec }^2{\theta }_2$
• C. ${\tan }^2\theta ={\tan }^2{\theta }_1+{\tan }^2{\theta }_2$
• D. ${\cot }^2\theta ={\cot }^2{\theta }_1+{\cot }^2{\theta }_2$

Answer 1

Answer: (D)

${\cot }^2\theta ={\cot }^2{\theta }_1+{\cot }^2{\theta }_2$

Note that $tan\theta =\frac{V}{H}$ ...............$\left(i\right)$

where $\theta$ is true value of dip . Here we assumed $\theta =\propto$

Now, angle of dips, ${\theta }_1$ and ${\theta }_2$ have different formula

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$tan{\theta }_1=\frac{V}{H\times cos\alpha }$ ................$\left(ii\right)$

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$tan{\theta }_2=\frac{V}{H\times sin\alpha }$ ................$\left(iii\right)$​

From eqn $\left(i\right),\left(ii\right),\left(iii\right)$, we can find value of $H/V$

which cotθ, which when calculate comes option D

$co{t}^2{\theta }_1+{cot}^2{\theta }_2$ = $\frac{H^2}{V^2}(co{s}^2\alpha +si{n}^2\alpha )=co{t}^2\theta$