Question

A dip circle shows an apparent dip of ${60}^0$ at a place where the true dip is ${45}^0$. If the dip circle is rotated by ${90}^0$, what will be the apparent dip?

• A. ${\cos }^{-1}\sqrt{\frac{2}{3}}$
• B. ${\tan }^{-1}\sqrt{\frac{2}{3}}$
• C. ${\sin }^{-1}\sqrt{\frac{2}{3}}$
• D. ${\cot }^{-1}\sqrt{\frac{2}{3}}$

Answer 1

The correct option is D ${\cot }^{-1}\sqrt{\frac{2}{3}}$

${\cot }^2\delta ={\cot }^2{\delta }_1+{\cot }^2{\delta }_2$
where ${\delta }_1$ and ${\delta }_2$ are dip angles in two vertical planes at right angles to each other and $\delta$ is the true angle of dip.
$\therefore {\cot }^2{\delta }_2={\cot }^2\delta -{\cot }^2{\delta }_1={\cot }^2{45}^{\circ }-{\cot }^2{60}^{\circ }=1-\frac{1}{3}=\frac{2}{3}$
${\delta }_2={\cot }^{-1}\sqrt{\frac{2}{3}}$