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At 45^(@)to the magnetic meridian theapparentis 60^(@) the true dip is |
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Answer» `TAN^(-1)SQRT(3)` `tan delta=(v)/(H)=(V)/(H cos 45^(@))=(V)/(H)xx(1)/(cos 45^(@))` but `tan 60^(@) =tan delta sqrt(2)` `tan delta =sqrt(3)//2therefore delta =tan^(-1) sqrt(3)//2` or `tan delta =sqrt(3)//2 therefore delta =tan^(-1) sqrt(3)//2` `rarr tan delta =tan delta cos alpha=sqrt(3)/(2)` |
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