A simple question about angles on spherical-geometry In the figure's sphere does:
$\widehat{AB} = \widehat{A'B'}$ or Not?
I mean the angles represented by arcs. (not the lengths).
In a book it said that they're not equal and they are:
 $\widehat{AB} = \widehat{A'B'} * \cos{\widehat{AA'}}$.
Is it right? If yes, How it is possible?

Example
In the above picture we know $AA' = BB' = 21^{o}$ (I don't write the degree in the other ones). and also A'B' (something like difference in longitudes of A and B) is: $A'B' = APB = A'PB' = \Delta\lambda = 66$
The book where I said the question solved like this:
The formula I said above say that the Arc $ADB = 66 *  \cos (21) = 61.61630814881532$ which is obviously not $66$.
If we think the green arc ($ACB$) is part of great circle then we have this formula:
$\cos (ADB) = \cos (PB) \cos(PA) + \sin(PB)\sin(PA)cos(APB)$
$\cos (ADB) = \cos (90-21) \cos(90-21) + \sin(90-21)\sin(90-21)cos(66)$
And from that we get $ADB = 61.123188817554769358$
so $66\neq 61.123188817554769358 \neq 61.61630814881532 $
So what does each of formulas actually measure?
 A: On the assumptions that


*

*$A'$ and $B'$ lie on the equator,

*$AA'$ and $BB'$ are longitudes,

*$A$ and $B$ lie on the same latitude,
the respective angles $\angle A'B'$ and $\angle AB$ subtended from the centers of the corresponding latitude lines are equal, but the angles subtended from the center of the sphere (i.e., the spherical distances) are not equal.
The formula $\widehat{AB} = \widehat{A'B'} \cos\widehat{AA'}$ appears to refer to a third quantity: length measured along the latitude segment $D$. This also clearly depends on the latitude.

Quantitatively, assume $A$ and $A'$ have longitude $0$, that $B$ and $B'$ have longitude $\theta = \angle A'B'$, that $A$ and $B$ have latitude $\phi = \angle AA'$, and that all four points lie on the unit sphere centered at the origin. In terms of these spherical coordinates, the four points have Cartesian coordinates:
\begin{align*}
  A &= (\cos\phi, 0, \sin\phi) & A' &= (1, 0, 0) \\
  B &= (\cos\theta \cos\phi, \sin\theta \cos\phi, \sin\phi) & B' &= (\cos\theta, \sin\theta, 0).
\end{align*}
The dot product gives
$$
\angle AB = \arccos[\cos\theta \cos^{2}\phi + \sin^{2}\phi]
= \arccos[\cos\theta + \sin^{2}\phi(1 - \cos\theta)]
$$
for the arc length of the spherical segment $C$.
The latitude $AB$, viewed as a circle in Euclidean $3$-space, has radius $\cos\phi$, and the formula $\widehat{AB} = \widehat{A'B'} \cos\widehat{AA'}$ gives length of the latitude segment $D$.
