Question

At an angle ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ to the magnetic meridian, the apparent dip is ${60}^{\circ }$. The true dip at the place is :

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${37}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

Angle between planes

$\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

${\theta }_A\to$ Apparent dip $={60}^{\circ }$

and ${\theta }_T\to$ True dip = ?

So, $\tan {\theta }_A=\frac{\tan {\theta }_T}{\cos \theta }$

$\Rightarrow \tan {60}^{\circ }=\frac{\tan {\theta }_T}{\left(\frac{1}{\sqrt{3}}\right)}$

$\tan {\theta }_T=\frac{1}{\sqrt{3}}\tan {60}^{\circ }=1$

$\Rightarrow {\theta }_T={45}^{\circ }$

Hence, option $\left(B\right)$ is the correct answer.