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.

Explanation

Here , `theta = 30^@`
Apparent value of dip ,` delta_1 = 45^@`
Actual value of dip ` delta=?`
If H is horizontal component of earth.s magnetic field in the magnetic meridian , then ` tan delta=(V)/(H)`
Let `H_1` be component of H at ` 30^@` to magnetic meridian , then ` tan delta_1 = (V)/(H_1) = (V)/(H cos theta ) = ( tan delta)/( cos theta )`
` or tan delta = tan delta_1 xx cos theta = tan 45^@ xx cos 30^@`
` = 1 xx (sqrt(3))/(2) = (1.732)/(2) = 0.866`