A magnetic needle suspended in a vertical plane at `30^@` from the magnetic meridian makes an angle of `45^@` with the horizontal. Find the true angle of dip.

Options

A. `45^(@)`
B. `41^(@)`
C. `31^(@)`
D. `60^(@)`

Answer

B

Explanation

In a vertical plane at `30^(@)` from the magnetic meridian, the horizontal component is ,

While vertical component is still V. Therefore , apparent dip will be given by ` tan theta =(V)/(H')=(V)/( H cos 30^(@))`
but `(V)/(H)= tan theta `( where ` theta `=true angle of dip )
`:. tan theta'=( tan theta )/( cos 30^(@))`
` :. theta =tan^(-1)[tan theta' cos 30^(@)]`
`= tan^(-1)[( tan 45^(@)) ( cos 30^@)]=41^(@)`