Dip angle in vertical plane at an angle ^ - 1 , ( 1 2 ) from magn… - Vidyadip

MCQ

Dip angle in vertical plane at an angle ${\cos ^{ - 1}}\,\left( {\frac{1}{{\sqrt 2 }}} \right)$ from magnetic meridian is $60^o$, then actual dip at that place

A
${\tan ^{ - 1}}\,\left( {\frac{{\sqrt 3 }}{2}} \right)$
B
${\tan ^{ - 1}}\,\left( {\frac{1}{{\sqrt 6 }}} \right)$
C
${\tan ^{ - 1}}\,\left( 1 \right)$
✓
${\tan ^{ - 1}}\,\left( {\sqrt {\frac{3}{2}} } \right)$

✓

Answer

Correct option: D.

${\tan ^{ - 1}}\,\left( {\sqrt {\frac{3}{2}} } \right)$

d
$\alpha=\cos ^{-1} \frac{1}{\sqrt{2}}, \Rightarrow \cos \alpha=\frac{1}{\sqrt{2}}, \theta^{\prime}=60^{\circ}$

$\tan \theta^{\prime}=\tan \theta / \cos \alpha$

so $\tan \theta=\tan \theta^{\prime} \cos \alpha$

$=\left(\tan 60^{\circ}\right) \frac{1}{\sqrt{2}}=\sqrt{3} \times \frac{1}{\sqrt{2}}$

$\theta=\tan ^{-1}(\sqrt{\frac{3}{2}})$

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