# ^Apparent dip

^**Apparent dip **

Let respectively be the vertical component, horizontal component & dip angle in a vertical plane inclined at some angle say α to magnetic meridian, then .

On dividing we wet tanδ^{/} = tanδ secα

From above relation we can write

- As sec α > 1, thus for any vertical plane inclined at some angle say α to magnetic meridian dip angle is greater than its value in magnetic meridian i.e. δ
^{/}> δ. - δ
^{/ }= 90^{0}if α = 90^{0}e. in a plane perpendicular to magnetic meriadian dip needle will be vertical.

In a similar way it can be proved that if δ_{1} and δ_{2} be the angles of dip observed in two vertical planes at right angles to each other and δ is the true angle of dip, then cot^{2} δ_{1} + cot^{2} δ_{2} = cot^{2 }δ.