Question
Horizontal component of earth’s magnetic field at a place is $\sqrt{3}$ times its vertical component. What is the value of angle of dip at that place?
✓
Answer
Given H $=\sqrt{3}\text{V}$
$\Rightarrow\tan\theta=\frac{\text{V}}{\text{H}}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{Angle of dip,}\theta=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)=30$
$\Rightarrow\tan\theta=\frac{\text{V}}{\text{H}}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{Angle of dip,}\theta=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)=30$
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