Question

prove that (i) tan 0=v/h and (ii) 12 = ∨2 + h2, where symbols have usual meanings.​

Answer 1

SOLUTION

TO PROVE

$\displaystyle \sf{(i) \: \: \tan \theta = \frac{V}{H} }$

$\sf{(ii) \: \: \: {I}^{2} = {V}^{2} + {H}^{2} }$

PROOF

Let

I = Total intensity of earth field

θ = The angle of dip

V = Vertical component of I

H = Horizontal component of I

Now resolving I along vertical horizontal direction we get

H = I cos θ

V = I sin θ

Then

$\displaystyle \sf{ \frac{V}{H} }$

$\displaystyle \sf{ = \frac{I \sin \theta}{ I\cos \theta } }$

$\displaystyle \sf{ = \tan \theta }$

$\boxed{ \: \: \displaystyle \sf{ \tan \theta = \frac{V}{H} } \: \: }$

Again

$\sf{ {V}^{2} + {H}^{2} }$

$\displaystyle \sf{ = {(I \sin \theta )}^{2} + { (I\cos \theta ) }^{2} }$

$\displaystyle \sf{ = {I}^{2}{ \sin }^{2} \theta + {I}^{2}{ \cos }^{2} \theta}$

$\displaystyle \sf{ = {I}^{2}({ \sin }^{2} \theta + { \cos }^{2} \theta})$

$\displaystyle \sf{ = {I}^{2} \times 1}$

$\displaystyle \sf{ = {I}^{2} }$

$\boxed{ \sf{ \: \: \: {I}^{2} = {V}^{2} + {H}^{2} } \: \: }$

Hence proved

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