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$\left(\theta \right)$ is given by ________

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

Answer 1

The correct option is D

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

Step 1: Given Data

The first apparent angle of dip $={\theta }_1$

The second apparent angle of dip $={\theta }_2$

The angle between the verticle planes $=90°$

Step 2: Formula Used

The dip angle is also referred to as the magnetic dip and is called the angle which is made by the earth’s magnetic field with the horizontal lines.

Let $V$ be the vertical component and $H$ be the horizontal component, then the angle of dip is given as

$\tan \theta =\frac{V}{H}$

Step 3: Solution

$\tan {\theta }_1=\frac{V}{H_1}$

$\tan {\theta }_2=\frac{V}{H_2}$

$H_1$ and $H_2$ are horizontal components between two planes at $90°$,

Thus,

$H^2=H_1^2+H_2^2$

$\Rightarrow {\left(\frac{V}{\tan \theta }\right)}^2={\left(\frac{V}{\tan {\theta }_1}\right)}^2+{\left(\frac{V}{\tan {\theta }_2}\right)}^2$
$\Rightarrow {(Vcot\theta )}^2={(Vcot{\theta }_1)}^2+{(Vcot{\theta }_2)}^2$

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

Therefore, the correct answer is option (D).