Question

If θ1​ and θ2​ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip θ is given by:


A \[ta{n^2}\theta = ta{n^2}{\theta _1} + ta{n^2}{\theta _2}\]

B \[co{t^2}\theta = co{t^2}{\theta _1} - co{t^2}{\theta _2}\]

C \[ta{n^2}\theta = ta{n^2}{\theta _1} - ta{n^2}{\theta _2}\]

D \[{\cot ^2}\theta = {\cot ^2}{\theta _1} + {\cot ^2}{\theta _2}\]

Answer 1

Hint –we will use the concept angle of dip here and we know that Angle of dip or magnetic dip is the angle that is made by the earth's magnetic field lines with the horizontal. ... When the horizontal component and the vertical component of the earth's magnetic field are equal, the angle of dip is equal to 45°.
In expression we can write \[\tan \theta = \dfrac{V}{H}\]
Where V vertical component and H is horizontal component

Formula used
 Angle of dip \[\tan \theta = \dfrac{V}{H}\]

Complete step by step solution
 We have given θ1​ and θ2​ be the apparent angles of dip observed in two vertical planes at right angles to each other
Since given observation is recorded in vertical plane so vertical component will be same for both the cases only changes in horizontal component
\[\therefore \tan {\theta _1} = \dfrac{V}{{{H_1}}}\]...................(1)
And \[\
  \tan {\theta _2} = \dfrac{V}{{{H_2}}} \\
  {H^2} = {H_1}^2 + {H_2}^2 \\
\ \].....................(2)
where \[{\theta _1}and{\theta _2}\] is the apparent angle of dip and \[{H_1}and{H_2}\] is the horizontal component and V is vertical component which is same for the cases
Now for true angle of dip we have
\[\
  \tan \theta = \dfrac{V}{H} \\
  H = V\cot \theta = \dfrac{1}{{\tan \theta }} \\
\ \].............................(3)
In equation (1) , (2) and (3) we are isolating horizontal component so,
\[H = \dfrac{V}{{\tan \theta }}\],\[{H_2} = \dfrac{V}{{\tan {\theta _2}}}\] and \[{H_1} = \dfrac{V}{{\tan {\theta _1}}}\]
Now substituting \[\cot \theta = \dfrac{1}{{\tan \theta }}\] in the above expression so,
\[H = V\cot \theta \]....................(4)
\[{H_1} = V\cot {\theta _1}\]...........................(5)
\[{H_2} = V\cot {\theta _2}\].........................(6)
Since it is given that plane are mutually perpendicular so we can write horizontal component as
\[{H^2} = {H_1}^2 + {H_2}^2\] (7)
Now putting the value of equation (4),(5) and (6) in equation (7)
\[V{\cot ^2}\theta = V{\cot ^2}{\theta _1} + V{\cot ^2}{\theta _2}\]
\[\therefore {\cot ^2}\theta = {\cot ^2}{\theta _1} + {\cot ^2}{\theta _2}\]

So option D is correct

Note:- Alternate method
If θ1​ and θ2 are Apparent angles of dip Let α be the angle which one of the plane make with the magnetic meridian
\[\tan {\theta _1} = \dfrac{{\tan \theta }}{{\cos \alpha }}\]
\[ \Rightarrow {\mkern 1mu} \tan {\theta _2} = \dfrac{{\tan \theta }}{{\cos (90 - \alpha )}} = \dfrac{{\tan \theta }}{{\sin \alpha }}\]
\[\because {\mkern 1mu} {\sin ^2}\alpha + {\cos ^2}\alpha = 1\]
\[\therefore {\cot ^2}\theta = {\cot ^2}{\theta _1} + {\cot ^2}{\theta _2}\]