In a certain place on the surface of earth, angle of dip in a vertical plane ‘P’ is ${30}^0$ . At the same place, angle of dip in another vertical plane ‘Q’ which is perpendicular to the plane. ‘P’, is  ${60}^0$. Then true angle of dip at that place is

• A

  ${\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

• B

  ${\tan }^{-1}\sqrt{\frac{7}{10}}$

• C

  ${\tan }^{-1}\left(\frac{\sqrt{3}}{4}\right)$

• D

  ${\tan }^{-1}\sqrt{\frac{3}{10}}$