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

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

angle of dip, $\tan \theta =\frac{vertical}{Horizontal}=\frac{v}{H}$
Case 1: $\tan {\theta }_1=\frac{v}{H_1}$
Case 2: $\tan {\theta }_2=\frac{v}{H_2}$
As given, $H_1$ and $H_2$ are $\perp$
Then, $\begin{array}{}{H}^2=H_1^2+H_2^2& \text{(1)}\end{array}$
putting above values in eq. (1)
we get,
${\left(\frac{v}{\tan \theta }\right)}^2={\left(\frac{v}{\tan {\theta }_1}\right)}^2+{\left(\frac{v}{\tan {\theta }_2}\right)}^2$

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