# Angle of declination and angle of dip

Question:

The plane of a dip circle is set in geographic meridian and the apparent dip is $$\delta_1$$. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is $$\delta_2$$. The declination $$\theta$$ at the plane is-

$$\theta= \tan^{-1}\left(\frac{\tan\delta_1}{\tan\delta_2}\right)$$

Attempt:

$$\tan\delta= \frac{\mathit{B_v}}{\mathit{B_h}}$$ where $$\delta$$ is angle of dip, $$\mathit{B_v}$$ and $$\mathit{B_h}$$ are vertical and horizontal components of Earth’s Magnetic Field at the location of dip circle.

For the two cases, $$\mathit{B_h}$$ would differ. My book says $$\mathit{B_{h_1}}$$ would be $$\mathit{B_h}\cos\theta$$ and $$\mathit{B_{h_2}}$$ would be $$\mathit{B_h}\cos(90-\theta)$$. Taking these in the equation, i can get the above mentioned answer.

What I am having trouble understanding is how these values of $$\mathit{B_h}$$ came to be? Could someone guide me with the help of a figure or something?

Thank you.

• Hi and welcome to physics.SE! Please do not post formulae as plain text, but use MathJax instead. – ACuriousMind Apr 30 '18 at 10:00

Angle of Dip $\delta$ is the angle in the vertical plane aligned with magnetic north (the magnetic meridian) between the local magnetic field and the horizontal.

Angle of Declination $\theta$ is the angle between the magnetic and geographic meridians, or the angle in the horizontal plane between magnetic north and true north. Source : TutorVista

In the magnetic meridian the vertical and horizontal components of the magnetic field are $B_v=B\sin\delta$ and $B_h=B\cos\delta$. The projection of the horizontal component $B_h$ onto the geographic meridian is $B_g=B_h\cos\theta$.

The plane perpendicular to the geographic meridian makes angle $180^{\circ}-(90^{\circ}+\theta)=90^{\circ}-\theta$ with the magnetic meridian. The projection of $B_h$ onto this plane is $B_p=B_h\cos(90^{\circ}-\theta)=B_h\sin\theta$.

The apparent angles of dip in the geographic meridian and the plane perpendicular to it are given by $$\tan\delta_1=\frac{B_v}{B_g}$$ $$\tan\delta_2=\frac{B_h}{B_p}$$ Therefore $$\frac{\tan\delta_1}{\tan\delta_2}=\frac{B_p}{B_g}=\frac{B_h\sin\theta}{B_h\cos\theta}=\tan\theta$$ $$\theta=\tan^{-1}(\frac{\tan\delta_1}{\tan\delta_2})$$