Real Life Applications of Geometry

It would be easier if I could post images here, but I don't. Wonder if any one can help me.

Let Earth's axis be the z-axis and the Sun is in the direction of x-axis. On summer solstice day, The line joining the centers of the Earth and Sum is normal to the Tropic of Cancer at $23.5^oN$ . Looking along the y-axis, we see the day-night line is at an angle of $23.5^o$ with the z-axis and Jaipur at $27^oN$ has more daylight than night.

Still looking along the y-axis, let the extra distance into daylight along the x-axis at $27^oN$ be $a$ . And if the Earth's radius is $R$ then:

$a = R\sin {27^o} \tan {23.5^o}$

The $27^oN$ circle has a radius $r = R \cos {27^o}$ . And if the angle of the arc in night darkness is $\theta$ , we note that:

$a = r \cos { \frac {\theta} {2} } = R \cos {27^o} \cos { \frac {\theta} {2} }$

$\Rightarrow \cos { \frac {\theta} {2} } = \tan {27^o} \tan {23.5^o} = 0.509525449 \times 0.434812375 = 0.221547971$

$\Rightarrow \frac {\theta}{2} = \cos^{-1} { 0.221547971} = 77.20003095^o\quad \Rightarrow \theta = 154.4000619^o$

Therefore, the percentage of daylight,

$x = 100 \left( 1 - \frac {\theta}{360^o} \right) = 100 \left( 1 - \frac {154.4000619^o }{360^o} \right) = 57.11109391\%$

The required answer $\lfloor 10x - 500 \rfloor = \lfloor 571.1109391 - 500 \rfloor = \boxed {71}$