Question

If ${\varphi }_1$ and ${\varphi }_2$ be the angles of dip observed in two vertical planes at right angles to each other and $\varphi$ be the true angle of dip, then

• A. $co{s}^2\varphi =co{s}^2{\varphi }_1+co{s}^2{\varphi }_2$
  
• B. $se{c}^2\varphi =se{c}^2{\varphi }_1+se{c}^2{\varphi }_2$
  
• C. $ta{n}^2\varphi =ta{n}^2{\varphi }_1+ta{n}^2{\varphi }_2$
  
• D. $co{t}^2\varphi =co{t}^2{\varphi }_1+co{t}^2{\varphi }_2$

Answer 1

The correct option is D $co{t}^2\varphi =co{t}^2{\varphi }_1+co{t}^2{\varphi }_2$

Let $\alpha$ be the angle which one of the planes make with the magnetic meridian the other plane makes an angle $({90}^{\circ }-\alpha )$ with it. The components of H in these planes will be $Hcos\alpha$ and $Hsin\alpha$ respectively. If ${\varphi }_1$ and ${\varphi }_2$ are the apparent dips in these two planes, then




$tan{\varphi }_1=\frac{V}{Hcos\alpha }i.e.cos\alpha =\frac{V}{Htan{\varphi }_1}$... (i)
$tan{\varphi }_2=\frac{V}{Hsin\alpha }i.e.sin\alpha =\frac{V}{Htan{\varphi }_2}$... (ii)
Squaring and adding (i) and (ii), we get
$co{s}^2\alpha +si{n}^2\alpha ={\left(\frac{V}{H}\right)}^2(\frac{1}{ta{n}^2{\varphi }_1}+\frac{1}{ta{n}^2{\varphi }_2})$
i.e. $1=\frac{V^2}{H^2}(co{t}^2{\varphi }_1+co{t}^2{\varphi }_2)$
or $\frac{H^2}{V^2}=co{t}^2{\varphi }_1+co{t}^2{\varphi }_2$ i.e. $co{t}^2\varphi =co{t}^2{\varphi }_1+co{t}^2{\varphi }_2$
This is the required result.