Question

If the apparent value of dip at a place in two perpendicular vertical planes are respectively ${30}^{\circ }$ and ${45}^{\circ }$, then the true angle of dip at that place is

• A. ${\cot }^{-1}\left(2\right)$
• B. ${\tan }^{-1}\left(2\right)$
• C. ${\cot }^{-1}\left(\sqrt{3}\right)$
• D. ${\cot }^{-1}\left(\sqrt{5}\right)$

Answer 1

The correct option is A ${\cot }^{-1}\left(2\right)$

Given:
${\delta }_1^{\prime }={30}^{\circ };{\delta }_2^{\prime }={45}^{\circ }$

Let,
$B_H=$ horizontal component of magnetic field, and
$B_V=$ vertical component of magnetic field


From the given diagram,

$\tan {\delta }_1^{\prime }=\frac{B_{V1}}{B_{H1}};\tan {\delta }_2^{\prime }=\frac{B_{V2}}{B_{H2}}$

As we know that vertical component of magnetic field due to all dip remains same.

$\therefore B_{V1}=B_{V2}=B_V$

So,
$B_{H1}=B_V\cot {\delta }_1^{\prime };B_{H2}=B_V\cot {\delta }_2^{\prime }$

Let $B_H$ and $B_V$ be the horizontal and vertical components of the magnetic field for the true dip.

$\therefore B_H=B_V\cot \delta$

Since, planes $\text{BC}$ and $\text{BD}$ are perpendicular.

$\therefore B_H^2=B_{H1}^2+B_{H2}^2$

$\Rightarrow {\cot }^2\delta ={\cot }^2{\delta }_1^{\prime }+{\cot }^2{\delta }_1^{\prime }$

$\Rightarrow {\cot }^2\delta ={\cot }^2{30}^{\circ }+{\cot }^2{45}^{\circ }=3+1$

$\therefore \delta ={\cot }^{-1}\left(2\right)$

Therefore, option $\left(A\right)$ is correct.