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The question is clear enough in my opinion but, I've found out that if you redraw this picture here:
You'll notice that the Sun is as big as the Earth. But if you scale up the Sun to it's real size compared to the Earth, you would realize that the Earth would covered in daylight ALL THE TIME. So what's going on here?
The sun is really far away. Thus its rays are essentially parallel at the earth's orbit. So, while the diagram you posted is clearly a bit off in terms of the relative size and distance between the sun and the earth, the parallel rays are about right.
The sun is about 100 times the size of the Earth (in diameter), and the distance from the sun to Earth is about 100 times the diameter of the sun. Below is an image showing the sun, Earth, and the distance between them to scale. It looks at first like nothing more than a black bar. This image in the original is 4000 pixels (the limit of my Pixlr Editor) by 60 pixels, and Earth is less than a pixel. To see it full size, click here or on the image, then when it comes up in your web browser click on it again to magnify it to full size. Use your scroll bar to see the sun at one end and Earth at the other. You will need to look really closely to see Earth.
But if you scale up the Sun to it's real size compared to the Earth,
Is pretty big. But then to be realistic in that manner you would also have to scale up the distance from the earth to the sun.
But you don't have to even do that, as there is a much simpler way of seeing how large the sun is from the perspective of a position on the earth.
If it's daytime and not cloudy you should see the sun up there somewhere (please don't keep looking…). It's not very big, about 32′ (varies slightly) which is just slightly bigger than 0.5°.
That's about the same as the moon (or else transits wouldn't be the interesting-looking eclipses that we get). Or about the size of the princess' thumb at arm's length in the story of the princess who wanted the moon for her birthday.
It's not enough to bathe most of the earth in the sun's rays. It's just about enough for almost half to be lit, almost half to be dark, and to have a period where a given point of the earth is illuminated by some, but not all, of the sun's visible disc. You can see this by the same look-at-the-sky experiment, by looking toward the horizon during sunrise or sunset.
Given the known average distance to the Sun, and the radii of Sun and Earth, the basic trigonometry is simple. If the Earth and Sun were exactly the same size, and there was no atmospheric refraction, then exactly half the planet, or 180 degrees, would be illuminated. But since the Sun is so much bigger, again assuming no atmospheric refraction, then exactly 180.522 degrees of the Earth would be illuminated. Calculating the additional illumination caused by atmospheric refraction of sunlight into the zone of darkness is fraught with difficulty because of variable atmospheric haze, temperature-controlled differences in air density, and cloud distribution. I'm guessing you could add a degree or two more if you were to include refracted twilight.
Because no matter how big the Sun is compared to the Earth, only half the Earth can be facing the Sun at any given time.
Try drawing a diagram where the Sun is much bigger with respect to the Earth. Make it ten times or a hundred times bigger or go all out and draw it to scale. Suppose it to be so big that it doesn't even look like a circle, but just a straight line of infinite length. Still, only half the Earth is facing it at any given time. Unless sunlight can penetrate the Earth, or somehow leaks around the edges, half the Earth will still be dark.
Update
Ahhh, I see where you're coming from now. Okay, true, if the Sun was infinitely large, or at least extremely large and extremely close, and if light left any given point on the surface of the Sun travelling in all directions, then yes, light rays could reach parts of the Earth not facing toward the Sun. They couldn't actually reach the point on the exact opposite side, but I guess close enough. I confess, I was thinking of light rays travelling from the Sun to the Earth in a direction parallel to a straight line from the center of the Earth to the center of the Sun.
I suppose that all light leaving the surface of the Sun is not travelling directly outward from the center, so to that extent, your model is plausible. Frankly I don't know enough about the physics to say how close that resembles reality. I'd think more light is travelling directly out from the center than at tangents, because the Sun is not a dark core with a thin glowing surface, but rather produces light from within. I have no idea what the relative numbers are though. But I guess the question is not the amount of light, but any light. So let's accept your model.
So yes, if the Sun was big enough and close enough, light from, say, the north pole of the Sun could hit a point on the "far side" of the Earth at a tangent. As in my diagram.
In real life, though, it's not THAT big or THAT close. Let's run the numbers.
Let $s$ be the distance from the Earth to the Sun, $r$ the radius of the Sun, and $A$ the angle along the curve of the Earth from the north pole. Then we have the triangle shown below, where $h$ is the height of the triangle. The radius of the Earth is very small compared to all the other numbers here, and so basically gets lost in the rounding. As $s$ is very large compared to $r$, $h$ is very close to $r$, so just to keep the geometry simple, let's assume $h=r$. (As $r \ge h$, calculating the real value for $h$ hurts your case.)
So geometry tells us that $r=s * \sin A$.
The Sun is about 93 million miles away. So for light rays in this scenario to reach 45 degrees past the Earth's north pole, we'd have:
$$ r = 93,000,000 * \sin(45) \\ r = 93,000,000 * \frac{\sqrt 2}{2} \\ r \approx 65,000,000 \\ $$
That is, the radius of the sun would have to be more than 65 million miles. In real life it's nowhere near that, it's more like 432,000.
So how far past the pole can the light really bleed?
$$ h \ge s * \sin A \\ \sin A \le h / s \\ \sin A \le \frac {432,000}{93,000,000} \\ \sin A \le .004645 \\ A \le 0.266 °\\ $$
That is, sunlight can "bleed" past the pole by about 1/4 of 1 degree.
