Question

A magnetic needle is free to rotate in a vertical plane which makes an angle of ${60}^{\circ }$ with the magnetic meridian. if the needle stays in a direction making an angle of $ta{n}^{-1}\left(\frac{2}{\sqrt{3}}\right)$ with the horizontal, what would be the dip at that place?

Answer 1

If ${\delta }_1$ and ${\delta }_2$ be the apparent dips shown by dip circle in the $2\perp r$ positions, the true dip $\delta$ is given by

${\cot }^2\delta ={\cot }^2{\delta }_1+{\cot }^2{\delta }_2$

$\Rightarrow {\cot }^2\delta ={\cot }^2{45}^o+{\cot }^2{53}^o$

$\Rightarrow {\cot }^2\delta =1.56\Rightarrow \delta =38.6\approx {39}^o$