Perimeter of tilted latitude of earth is there any formula for the perimeter (or length) of part of latitude A degree north of earth, which gets sun light, when tilted towards sun by an angle B degrees??
 A: If the Sun is at declination $\delta$, then at latitude $\beta$, the semi-diurnal arc $\Delta$ is given by
$$
\cos(\Delta)=-\tan(\delta)\tan(\beta)\tag{1}
$$
Assuming that the Earth is a sphere of radius $R$, we can adjust $(1)$ to get the length of the portion of a parallel of latitude along which the Sun is above the horizon, we get
$$
2R\cos(\beta)\arccos(-\tan(\delta)\tan(\beta))\tag{2}
$$
The arccosine needs to be taken in radians.
For example, at mid-summer, the declination of the Sun is approximately $23^\circ27^\prime$ and the latitude of Bangalore is $12^\circ59^\prime$, assuming the radius of the Earth is $6371$ km, we get a length of $20747$ km.

Computation of Semi-Diurnal Arc
Since declination, latitude, and altitude are all measured from a point $90^\circ$ from the pole, the triangle of relevant angles on the sphere looks like
$\hspace{3.5 cm}$
where co-$\beta$ is the complement of the latitude, co-$\delta$ is the complement of the declination, $\Delta\lambda$ is the difference in hour angles or longitude, and co-alt is the complement of the altitude. The Spherical Law of Cosines says
$$
\sin(\text{alt})=\sin(\delta)\sin(\beta)+\cos(\delta)\cos(\beta)\cos(\Delta\lambda)\tag{3}
$$
When the Sun is on the horizon, it has an altitude of $0$. Thus, we get
$$
\cos(\Delta\lambda)=-\tan(\delta)\tan(\beta)\tag{4}
$$
which is the formula in $(1)$.
