A) \[{{\tan }^{-1}}\frac{1}{2}\]
B) \[{{\tan }^{-1}}(2)\]
C) \[{{\tan }^{-1}}\left( \frac{2}{3} \right)\]
D) None of these
Correct Answer: B
Solution :
\[\tan {{\varphi }^{'}}=\frac{\tan \varphi }{\cos \beta };\]where \[{{\varphi }^{'}}\]=Apparent angle of dip, \[\varphi \]= True angle of dip, \[\beta \]=Angle made by vertical plane with magnetic meridian. \[\Rightarrow \tan {{\varphi }^{'}}=\frac{\tan {{60}^{o}}}{\cos {{30}^{o}}}=2\Rightarrow {{\varphi }^{'}}={{\tan }^{-1}}\left( 2 \right)\]
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