A) \[{{\tan }^{-1}}\left( \frac{1}{2} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{1}{3} \right)\]
C) \[{{\tan }^{-1}}\left( \frac{1}{6} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{3}{7} \right)\]
E) \[{{\tan }^{-1}}\left( \frac{4}{3} \right)\]
Correct Answer: D
Solution :
At\[60{}^\circ \]to the magnetic meridian, the effective horizontal component of the earths magnetic field is \[B{{}_{H}}={{B}_{H}}\cos 60{}^\circ =\frac{1}{2}{{B}_{H}}\] The apparent dip \[(\delta )\] is given by \[\tan \delta =\frac{{{B}_{V}}}{{{B}_{H}}}=\frac{2{{B}_{V}}}{{{B}_{H}}}=2\tan \delta \] where, \[\delta =\text{true}\,\,\text{dip}\] Thus, \[\tan 45{}^\circ =2\tan \delta \] \[\Rightarrow \] \[\delta ={{\tan }^{-1}}\left( \frac{1}{2} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
