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-

Answer:

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

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.
 A: Start with some definitions :
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 $$\frac{\tan\delta_1}{\tan\delta_2}=\frac{B_p}{B_g}=\frac{B_h\sin\theta}{B_h\cos\theta}=\tan\theta$$
$$\theta=\tan^{-1}(\frac{\tan\delta_1}{\tan\delta_2})$$
