Question

at a place if Bh:BV=1:√3,what is the angle of dip at that place?​

Answer 1

$\tanθ = \frac{bh}{bv}$

$\tan θ= \frac{bh}{bv} = \frac{1}{ \sqrt{3} }$

$\tanθ= \frac{1}{ \sqrt{3} }$

$\huge\purple{\boxed{ θ= 30°}}$

$\huge\green{ツ}$