Question

If actual angle of dip is $\theta$ and $\theta'$ is the angle of dip in a plane at an angle $\alpha$ from the magnetic meridian, then $\dfrac{\tan \theta'}{\tan \theta}$ is

• A. $\sec \alpha$
• B. $\cos \alpha$
• C. $\cosec \alpha$
• D. $\cot \alpha$

Hint:

Remember: $\tan \theta' = \dfrac{\tan \theta}{\cos \alpha}$ for apparent dip. If $\alpha = 0$, then $\theta' = \theta$.

Answer 1

The Correct Option is A

Step 1: Recall concept of apparent dip.
If actual dip is $\theta$, and the dip needle makes an apparent dip $\theta'$ in a plane inclined at angle $\alpha$ to the magnetic meridian, then the relation is: \[ \tan \theta' = \frac{\tan \theta}{\cos \alpha}. \]
Step 2: Take ratio.
\[ \frac{\tan \theta'}{\tan \theta} = \frac{1}{\cos \alpha} = \sec \alpha. \]
Step 3: Conclusion.
Hence, the correct answer is (A) $\sec \alpha$.