question_answer

If \[{{\varphi }_{1}}\] and \[{{\varphi }_{2}}\] be the angles of dip observed in two vertical planes at right angles to each other and f be the true angle of dip, then

A)            \[{{\cos }^{2}}\varphi ={{\cos }^{2}}{{\varphi }_{1}}+{{\cos }^{2}}{{\varphi }_{2}}\]

B)            \[{{\sec }^{2}}\varphi ={{\sec }^{2}}{{\varphi }_{1}}+{{\sec }^{2}}{{\varphi }_{2}}\]

C)            \[{{\tan }^{2}}\varphi ={{\tan }^{2}}{{\varphi }_{1}}+{{\tan }^{2}}{{\varphi }_{2}}\]

D)            \[{{\cot }^{2}}\varphi ={{\cot }^{2}}{{\varphi }_{1}}+{{\cot }^{2}}{{\varphi }_{2}}\]

Correct Answer: D

Solution :

                   Let a be the angle which one of the planes make with the magnetic meridian the other plane makes an angle \[({{90}^{o}}-\alpha )\]with it. The components of H in these planes will be \[H\cos \alpha \]and \[H\sin \alpha \] respectively. If \[{{\varphi }_{1}}\]and \[{{\varphi }_{2}}\]are the  apparent dips in these two planes, then           \[\tan {{\varphi }_{1}}=\frac{V}{H\cos \alpha }\]  i.e. \[\cos \alpha =\frac{V}{H\tan {{\varphi }_{1}}}\] ..... (i)                    \[\tan {{\varphi }_{2}}=\frac{V}{H\sin \alpha }\] i.e. \[\sin \alpha =\frac{V}{H\tan {{\varphi }_{2}}}\]   ..... (ii)                    Squaring and adding (i) and (ii), we get                    \[{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha ={{\left( \frac{V}{H} \right)}^{2}}\left( \frac{1}{{{\tan }^{2}}{{\varphi }_{1}}}+\frac{1}{{{\tan }^{2}}{{\varphi }_{2}}} \right)\]                    i.e. \[1=\frac{{{V}^{2}}}{{{H}^{2}}}\left( {{\cot }^{2}}{{\varphi }_{1}}+{{\cot }^{2}}{{\varphi }_{2}} \right)\]                    or \[\frac{{{H}^{2}}}{{{V}^{2}}}={{\cot }^{2}}{{\varphi }_{1}}+{{\cot }^{2}}{{\varphi }_{2}}\]i.e. \[{{\cot }^{2}}\varphi ={{\cot }^{2}}{{\varphi }_{1}}+{{\cot }^{2}}{{\varphi }_{2}}\]            This is the required result.