A) \[{{\tan }^{-1}}\left( \sqrt{\frac{3}{2}} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]
C) \[{{\tan }^{-1}}\left( \sqrt{3} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{2}{\sqrt{3}} \right)\]
Correct Answer: B
Solution :
Let the real dip be \[\phi ,\] then \[\tan \phi =\frac{{{B}_{V}}}{{{B}_{H}}}\] For apparent dip,- ? \[\tan \phi =\frac{{{B}_{V}}}{{{B}_{H}}\cos \beta }=\frac{{{B}_{V}}}{{{B}_{H}}\cos 30{}^\circ }\] \[=\frac{2{{B}_{V}}}{\sqrt{3}{{B}_{H}}}\] or \[\tan 45{}^\circ =\frac{2}{\sqrt{3}}\tan \phi \] \[\phi ={{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]
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