Question

If apparent dip angles in two vertical planes situated at complementary angles from magnetic meridian are ${\cot }^{-1}\left(3/5\right)$, and ${\cot }^{-1}\left(4/5\right)$, then true dip angle at that place is

• A. ${\cot }^{-1}\left(5\right)$
• B. ${\cot }^{-1}\left(1/5\right)$
• C. ${45}^o$
• D. ${90}^o$

Answer 1

The correct option is C ${45}^o$

We know that

$\begin{array}{c}\tan {\theta }_1=\frac{v}{h_1}\\ \tan {\theta }_2=\frac{v}{H_2}\\ H_1=v\cot {\theta }_1\\ H_2=\cot {\theta }_2\\ \tan \theta =\frac{v}{H}\\ H=v\cot \theta \end{array}$

where ,

$v=$vertical component of magnetic field intensity

$H=$horizontal component of magnetic field intensity

As$H_1\&H_2$ are horizontal component in two planes at ${90}^{\circ }$to each other

$\begin{array}{c}{H}^2={H_1}^2+{H^2}_2\\ {\left(v\cot \theta \right)}^2={\left(v\cot {\theta }_1\right)}^2+\left(v_2\cot {\theta }_2\right)\\ {\cot }^2\theta ={\cot }^2{\theta }_1+{\cot }^2{\theta }_2\end{array}$

where

$\theta =$ true angle of dip

${\theta }_1\&{\theta }_2=$ two apparent angle of dip

$\begin{array}{c}{\theta }_1={\cot }^{-1}\left(\frac{3}{5}\right)\\ {\theta }_2={\cot }^{-1}\left(\frac{4}{5}\right)\\ \cot {\theta }_1=\frac{3}{5}\\ \cot {\theta }_2=-\frac{4}{5}\\ {\cot }^2\theta ={\left(\frac{3}{5}\right)}^2+{\left(\frac{4}{5}\right)}^2\\ =\frac{9}{25}+\frac{16}{25}\\ {\cot }^2\theta =1\\ \theta ={\cot }^{-1}\left(1\right)\\ \theta ={45}^{\circ }\end{array}$

$\left(C\right)$is the correct answer