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