Question

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

• A. ${\tan }^2\theta ={\tan }^2{\theta }_1-{\tan }^2\theta 2$
• B. ${\cot }^2\theta ={\cot }^2{\theta }_1+{\cot }^2\theta 2$
• 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

The correct option is B ${\cot }^2\theta ={\cot }^2{\theta }_1+{\cot }^2\theta 2$

Note that $\tan \theta =\frac{V}{H}$ ...(1)
where $\theta$ is true value of dip . Here we assumed $\theta =\propto$
Now, angle of dips, ${\theta }_1$ and ${\theta }_2$ have different formula
$\tan {\theta }_1=\frac{V}{H\times cos\propto }$ ...(2)
$\tan {\theta }_2=\frac{V}{H\times sin\propto }$ ...(3)
From eqn (1), (2), (3), we can find value of H/V
which $\cot \theta$, which when calculate comes option B
${\cot }^2{\theta }_1+{\cot }^2{\theta }_2=\frac{H_2}{V^2}({\cos }^2\propto +{\sin }^2\propto )={\cot }^2\theta$
${\cot }^2\theta ={\cot }^2{\theta }_1+{\cot }^2{\theta }_2$