Question

If ${\delta }_1$ and ${\delta }_2$ are the angles of dip observed in two vertical planes at right angles to each other and $\delta$ is the true value of dip then :

• A. $ta{n}^2\delta =ta{n}^2{\delta }_1+ta{n}^2{\delta }_2$
• B. $co{t}^2\delta =co{t}^2{\delta }_1+co{t}^2{\delta }_2$
• C. $ta{n}^2\delta =\frac{ta{n}^2{\delta }_1+ta{n}^2{\delta }_2}{ta{n}^2{\delta }_{1}ta{n}^2{\delta }_2}$
• D. $co{t}^2\delta =1+co{t}^2{\delta }_{1}co{s}^2{\delta }_2$

Answer 1

Answer: (B)

$co{t}^2\delta =co{t}^2{\delta }_1+co{t}^2{\delta }_2$

Note that $\tan \delta =\frac{V}{H}$ ............(1)
where $\delta$ is true value of dip. Here we assumed del equal to alpha.
Now,Angle of dips, ${\delta }_1$ and ${\delta }_2$ have different formluae.
$\tan {\delta }_1=\frac{V}{H\times \cos \alpha }$ ...........(2)
$\tan {\delta }_2=\frac{V}{H\times \sin \alpha }$............(3)
Using equations 1,2 and 3, we can find value of $\frac{H}{V},$ which is $\cot \delta$, which when calculated comes option B.
$co{t}^2{\delta }_1+co{t}^2{\delta }_2=\frac{H^2}{V^2}(co{s}^2\alpha +si{n}^2\alpha )=co{t}^2\delta$