Where on Earth (75% of the time in the dark)

[rede96]

...would I have to be so I spend 75% of the time in the dark as the Earth spins?

Imagine the sun shines on the Earth in such a way that it casts a shadow that is at 60 degrees to the axis of the spin of the Earth (Assume the Earth's axis is 0 degrees) as shown in my diagram below.

I am trying to work out what 'latitude' (shown as P) I have to be at so I spend 75% of my time in the dark and 25% of my time in sun light as the Earth spins about its axis. I am not looking for the actual latitude, but how to calculate d2 (diameter) and h2 (height) from below.

Can anyone help? I've been at this for ages!

 

[Nugatory]

How much trigonometry have you studied?
(Knowing this will help us give you a better answer)

 

[rede96]

How much trigonometry have you studied?

I know the basics, SOH, CAH, TOA. So can work most simple trig problems out.

 

[rede96]

If I look in the downward on the 'Earth' at the circle drawn out by the chord d2, then I know that 25% of the circumference of that circle needs to be in the light and 75% in the dark, which is 270 degrees and 90 degrees respectively. So I can make a segment that splits the circumference of a circle at 0 and 90 degrees, and thus work out the chord of that segment, and the arc which the shadow casts on the earth. But to do all that I would need to know d2 and I just can't figure out how to calculate that :(

 

[rede96]

A lot of trial and error but I managed to solve this. Below are the un-simplified equations. If anyone could help simplify them that would be great :)

From the diagram above:

h2 = sin((cos(((1-(cos(θ/2)^2))*360)/2)+1)/2*θ)*(d/2)

d2 = cos((cos(((1-(cos(θ/2)^2))*360)/2)+1)/2*θ)*d/2)*2

Where θ is the angle of the shadow (60° in the example I gave) and d is the diameter of the Earth (I called it d1 in the above diagram)