This geometry occurs in the definition of the angle of solar declination. Planes $X$ and $Y$ intersect along line $AF$. The angle between the two planes is defined as $a_0$ which is equivalently the angle $\angle BAC$. Now the side $AB$ is rotated to a new position denoted by $AD$ by an angle $p$ (or $\varphi$, but $p$ is what's in the image). The new angle $a_1 = \angle DAE$ is a function of $a_0$ and $p$. What is the expression for $a_1$ in terms of $a_0$ and $p$?

Note that the angle $p = \angle DAB$, all of which points lie in plane $X$. The triangles $ABC$ and $ADE$ are drawn to show the geometry but are not required to define the problem. This problem occurs between Earth and the Sun as Earth rotates around the Sun. The angle $p$ is $0$ or $180^\circ$ at solstice. The angle $a_0$ is the maximum declination angle of $23.45^\circ$.

An approximate answer appears to be $\sin(a_1) = \sin(a_0) \sin(90-p)$. How can this be proven?

@Peter gave a rigorous vector based proof. The following is a geometry based proof. Please see if the logic is correct.enter image description here. The geometry is constructed by choosing vectors AC and AE with unit length. Extended line CE meets AF at point Q. Once the triangles QAB and QBC are constructed as shown, the lengths of remaining sides are established using length/sin rule for a triangle. Observe that the rotation angle "q" instead of "p" is used to determine the lengths QE and EC. Then using similarity rule for triangles the ration of sides DE:BC is equated to QE:QC. The trigonometric formula Cos(A-B)=CosACosB + SinASinB allows L1/(L1+L2) to be reduced to Cos(q).

If the rotation angle "p" is referenced instead of "q" then the proof is repeated with vectors AB and AD with unit length. In this case the lengths of sides BC and DE becomes tan(a0) and tan(a1) which will results in tan(a1)=tan(a0)*cos(p).

Does this make sense? @Peter any comment?

Since angle p and q are not identical and they are linked to each other, the results makes sense.

9/20/2013 Simplest Geometry-based Proof:

Please refer to the following figure. Objective is to link angles a0 and a1 to the rotation angle either "p" or "q". The angle between the two planes X&Y is defined as a0. Hence the plane ABC is normal to both planes X and Y. By rotating the plane ABC to a new plane ADE by amount "p" (defined by BAD in plane-X) or by amount "q" (defined by CAE in plane-Y) the new angle "a1" is obtained. Note that the plane BCED is generated by drawing it parallel to line AF while keeping it normal to plane-X. This generates the rectangle BCED. Once BCED is understood to be a rectangle the derivation of the relationship is simple as shown in the figure. enter image description here

In the definition of solar declination angle, plane-X encompasses the orbit of earth around the sun and is referred to as ecliptic plane. Plane-Y is the equatorial plane which is normal to the axis of rotation of the earth. In one year as earth orbits the sun, the angle "p" evolves from 0 through 360 degrees at a constant rate of 360/365.25 degrees per day. Note that angle "p" is zero when the earth axis is tilted towards the sun on June 21 (solstice) at a maximum of 23.45 degrees (=a0). It is of interest to estimate a1 vs. p as earth orbits the sun. In the solar literature the relation ship sin(a1)=sin(a0)*cos(p) is used where as it should be tan(a1)=tan(a0)*cos(p). The angle "q" is approximately same as "p" but it will evolve not at constant rate of 360/365.25 degree/day. Is this observation correct?


Previous answer can be viewed in revision history, but is based on an incorrect interpretation of $p$.

Without loss of generality we can take $|AB| = |AC| = 1$ and place all of the other points on a unit sphere centred at $A$. Then since the planes all pass through $A$ their intersections with the sphere are great circles, and we have a spherical polygon $BCED$. Since $a_0$ is the angle between the planes defining $BD$ and $CE$, we find that $BC$ is perpendicular to those planes. Similarly, we find the $p$ is the length of the perpendicular between $BC$ and $DE$:

Spherical polygon with known lengths and angles

There is some obvious symmetry: the line of length $p$ bisects $BC$ and $DE$ (rather than being equal to $BD$ and $CE$).

If we were looking at distance along a line of constant latitude then we would have $$*\;a_1 = a_0 \cos p$$ but we're looking at great circle distance instead.

Call the midpoint of $BC$ $G$ and the midpoint of $DE$ $H$. Then let $\angle GHC = \alpha$ and $\angle GCH = \beta$. (Note: $\alpha + \beta > \frac{\pi}{2}$ since this is spherical geometry, not Euclidean).

Spherical polygon with triangulation

We have all of the angles and two of the sides of $GCH$, and the one remaining is opposite a right angle, so the spherical law of cosines gives us $$\cos |CH| = \cos p \cos \frac{a_0}{2}$$ and the spherical law of sines gives us $$\sin |CH| = \frac{\sin p}{\sin \beta} = \frac{\sin \frac{a_0}{2}}{\sin \alpha}$$

Now let's take the other triangle. We're have a side and its two angles, so we want to use a cotangent four-part formula: $$\cos |CH| \cos (\frac{\pi}{2} - \beta) = \cot \frac{a_1}{2} \sin |CH| - \cot (\frac{\pi}{2} - \alpha) \sin (\frac{\pi}{2} - \beta)$$

Eliminating those $\frac{\pi}{2}$, $$\cos |CH| \sin \beta = \cot \frac{a_1}{2} \sin |CH| - \tan \alpha \cos \beta$$

But that leaves us with three variables to eliminate, and non-linear equations to do it with.

  • $\begingroup$ Peter, thanks for the derivation. Seemingly simple geometry requires lot of vector products to get to the exact answer. Thanks for the steps. Yes, I have corrected my approximate answer where sin(p) should have been sin (90-p) which is cos(p). Your derivation confirms that the answer is indeed exact. Is there a geometry based proof in which properties of a triangle [for example, length-of-a-side/sin(angle) rule can be used to derive the same answer)? $\endgroup$ – Sri Sep 17 '13 at 16:04
  • $\begingroup$ @Sri, there's probably a more elegant geometric approach, but I'm not a geometer and I attack geometry problems by brute force. Although now that I think about it, if you're familiar with spherical geometry then by simply assuming that $|B| = |C|$ you can reduce it to a simple spherical triangle problem. $\endgroup$ – Peter Taylor Sep 17 '13 at 16:11
  • $\begingroup$ Hmm, maybe not quite as simple as I thought. $\endgroup$ – Peter Taylor Sep 17 '13 at 16:24
  • $\begingroup$ please review a geometry based proof. The reference plane on which the rotation angle "p" is defined seems to affect the final relationship. It is either sin(a1)=sin(a0)*cos(p) or tan(a1)=tan(a0)*cos(p) depending on the choice of the plane. $\endgroup$ – Sri Sep 18 '13 at 16:20
  • $\begingroup$ a simpler geometric argument and proof is added 9/20/2013. Any comments? $\endgroup$ – Sri Sep 20 '13 at 14:37

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