question_answer

The plane of a dip circle is set in the geographic meridian and the apparent dip is\[{{\delta }_{1}}\]. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is \[{{\delta }_{2}}\].The decimation\[\theta \]at the plane is

A) \[\theta ={{\tan }^{-1}}(\tan {{\delta }_{1}}\tan {{\delta }_{2}})\]

B) \[\theta ={{\tan }^{-1}}(\tan {{\delta }_{1}}+\tan {{\delta }_{2}})\]

C) \[\theta ={{\tan }^{-1}}\left( \frac{\tan {{\delta }_{1}}}{\tan {{\delta }_{2}}} \right)\]

D) \[\theta ={{\tan }^{-1}}(\tan {{\delta }_{1}}-\tan {{\delta }_{2}})\]

Correct Answer: C

Solution :

Here,\[\tan {{\delta }_{1}}=\frac{v}{H\cos \theta }\] \[\tan {{\delta }_{2}}=\frac{v}{H\cos ({{90}^{o}}-\theta )}=\frac{v}{H\sin \theta }\]                 \[\frac{\tan {{\delta }_{1}}}{\tan {{\delta }_{2}}}=\frac{\sin \theta }{\cos \theta }=\tan \theta \] or            \[\theta ={{\tan }^{-1}}\left( \frac{\tan {{\delta }_{1}}}{\tan {{\delta }_{2}}} \right)\]