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At 45^(@) to the magnetic meridian the ...

At `45^(@)` to the magnetic meridian the apparent is `60^(@)` the true dip is

A

`tan^(-1)sqrt(3)`

B

`tan^(-1)""(1)/(sqrt(3))`

C

`tan^(-1)sqrt(3/2)`

D

`tan^(-1)sqrt(1/6)`

Text Solution

Verified by Experts

The correct Answer is:
C
Apparent dip `delta` is given by
`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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