# 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.
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 $$\frac{\tan\delta_1}{\tan\delta_2}=\frac{B_p}{B_g}=\frac{B_h\sin\theta}{B_h\cos\theta}=\tan\theta$$ $$\theta=\tan^{-1}(\frac{\tan\delta_1}{\tan\delta_2})$$