Question

At ${45}^{°}$ of the magnetic meridian angle of dip is $30°$ then find the angle of dip in a vertical plane at $45°$ ?

• A. ${\tan }^{-1}\left(\frac{1}{\sqrt{6}}\right)$
• B. ${\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
• C. ${\tan }^{-1}\left(\frac{1}{\sqrt{4}}\right)$
• D. ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Answer 1

The correct option is A

${\tan }^{-1}\left(\frac{1}{\sqrt{6}}\right)$

Step 1. Given data

At an angle of $45°$, the angle of dip, ${\theta }_1$$=30°$

Step 2. Finding the ratio of the vertical component, $V$ and the horizontal component, $H$ of earth's magnetic field at the meridian.

By using the formula of the angle of dip,

$\tan {\theta }_1=\frac{V}{H\cos 45°}$

$\tan 30°=\frac{V\times \sqrt{2}}{H\times 1}$

$\frac{1}{\sqrt{3}}=\frac{V\sqrt{2}}{H}$

$\frac{V}{H}=\frac{1}{\sqrt{6}}$

Step 3. Finding the angle of dip, ${\theta }_2$ a vertical plane.

By using the formula,

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

$\tan {\theta }_2=\frac{1}{\sqrt{6}}$

${\theta }_2={\tan }^{-1}\left(\frac{1}{\sqrt{6}}\right)$

Hence, the correct option is $\left(A\right)$