Question

If ${\theta }_1$ and ${\theta }_2$ be the apparent of dip observed in two vertical planes at right angles to each other on either side of magnetic meridian, then the true angel of dip $\theta$ is given by

• A. $ta{n}^2\theta =ta{n}^2{\theta }_1+ta{n}^2{\theta }_2$
• B. $co{t}^2\theta =co{t}^2{\theta }_1-co{t}^2{\theta }_2$
• C. $ta{n}^2\theta =ta{n}^2{\theta }_1-ta{n}^2{\theta }_2$
• D. $co{t}^2\theta =co{t}^2{\theta }_1+co{t}^2{\theta }_2$

Answer 1

The correct option is D $co{t}^2\theta =co{t}^2{\theta }_1+co{t}^2{\theta }_2$

$Solution:$ Note that tan $\theta$ = $\frac{V}{H}$ $.......\left(1\right)$

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

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

tan ${\theta }_1$ = $\frac{V}{H\times cos\alpha }$ $.......\left(2\right)$

tan${\theta }_2$ =$\frac{V}{H\times sin\alpha }$ $.......\left(3\right)$

From equation (1),(2),(3),we get;

cot${}^2$${\theta }_1$ $+$ cot${}^2$${\theta }_2$ $=$ $\frac{H_2}{V^2}$(cos${}^2$${\alpha }_1$ $+$ sin${}^2$${\alpha }_2$) $=$ cot${}^2$$\theta$

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

$So,the$ $correct$ $option:D$