Looking at the sun during an eclispe

[BMAD]

Three students are outside looking at the sun during an eclispe (don't worry, they are taking precautions). Assume the students are the same height. The three students are collinear but an unknown distance apart. The first student is able to see the sun tilting up their head at a 30 degree angle. The second student needs to tilt their head at a 45 degree angle in order to see the sun. Now though we don't know how far apart the three students are from each other, we do know that second student is perfectly in the middle between the first and third. What angle must the third student use to see the sun?

[BMAD]

15 degrees?

the answer is approx (if not exactly) 75 degrees.

[MathewStephenKtm]

I think the third student must use 90 degree angle to see the sun.

[bonanova]

Imagine the sun is a lamp on a post 1 unit above the ground.
The distances from the lamp of the two end observers are

cot 45^o = 1.000 units and cot 30^o = 1.732 units.

The distance of the middle observer from the lamp post is thus 1.366 units.

His angle is cot^-1(1.366) = about 36.206^o.

Now the fact we did this with a streetlamp instead of the sun allowed us to greatly reduce the scale of the problem, and place the observers on a (very nearly) planar stage.

If we did this with the sun, the two end observers would be on different continents, the middle observer would be on a cruise ship basking in the sun, wondering why it got cold and dark momentarily, and forgetting to look. Moreover, we'd have to consider curvature of the earth changing the angle of inclination.

All of those things would make a wonderfully complex puzzle.

[dark_magician_92]

either 68.9 degrees or 54.89 degrees

[BMAD]

Imagine the sun is a lamp on a post 1 unit above the ground.

The distances from the lamp of the two end observers are

cot 45^o = 1.000 units and cot 30^o = 1.732 units.

The distance of the middle observer from the lamp post is thus 1.366 units.

His angle is cot^-1(1.366) = about 36.206^o.

Now the fact we did this with a streetlamp instead of the sun allowed us to greatly reduce the scale of the problem, and place the observers on a (very nearly) planar stage.

If we did this with the sun, the two end observers would be on different continents, the middle observer would be on a cruise ship basking in the sun, wondering why it got cold and dark momentarily, and forgetting to look. Moreover, we'd have to consider curvature of the earth changing the angle of inclination.

All of those things would make a wonderfully complex puzzle.

the first student used a 30 degree angle and the second student used a 45 degree angle. The op did not give the third person's angle.

[bonanova]

Imagine the sun is a lamp on a post 1 unit above the ground.

The distances from the lamp of the two end observers are

cot 45^o = 1.000 units and cot 30^o = 1.732 units.

The distance of the middle observer from the lamp post is thus 1.366 units.

His angle is cot^-1(1.366) = about 36.206^o.

Now the fact we did this with a streetlamp instead of the sun allowed us to greatly reduce the scale of the problem, and place the observers on a (very nearly) planar stage.

If we did this with the sun, the two end observers would be on different continents, the middle observer would be on a cruise ship basking in the sun, wondering why it got cold and dark momentarily, and forgetting to look. Moreover, we'd have to consider curvature of the earth changing the angle of inclination.

All of those things would make a wonderfully complex puzzle.

the first student used a 30 degree angle and the second student used a 45 degree angle. The op did not give the third person's angle.

I guess I don't understand. Was that not the question?

[BMAD]

Imagine the sun is a lamp on a post 1 unit above the ground.

The distances from the lamp of the two end observers are

cot 45^o = 1.000 units and cot 30^o = 1.732 units.

The distance of the middle observer from the lamp post is thus 1.366 units.

His angle is cot^-1(1.366) = about 36.206^o.

Now the fact we did this with a streetlamp instead of the sun allowed us to greatly reduce the scale of the problem, and place the observers on a (very nearly) planar stage.

If we did this with the sun, the two end observers would be on different continents, the middle observer would be on a cruise ship basking in the sun, wondering why it got cold and dark momentarily, and forgetting to look. Moreover, we'd have to consider curvature of the earth changing the angle of inclination.

All of those things would make a wonderfully complex puzzle.

the first student used a 30 degree angle and the second student used a 45 degree angle. The op did not give the third person's angle.

I guess I don't understand. Was that not the question?

maybe i am confused. but the way i read your solution is that you assumed the first person was 30 degrees, and the third person was 45 degrees and then found the second person.

The op gave the first person's angle and the second person's angle and asked for the thirds.

[Rob_Gandy]

the answer is approx (if not exactly) 75 degrees.

[BMAD]

the answer is approx (if not exactly) 75 degrees.

That is 1 of the two possible answers.

[Rob_Gandy]

15 degrees?

[BMAD]

15 degrees?

No but I think you have the right idea.

[bonanova]

I misread the order of the students.

If the light is a unit distance high, the middle observer is standing at a unit distance away.
The other two are approximately .732 units closer to and farther from the light.

cot^-1(0.268..., 1, 1.732...) = 75^o, 45^o and 30^o respectively.

However .... That is not the correct answer.

As I alluded in my first post, if we are talking about the Sun, then the unit height of the light is 1 Astronomical Unit, or 149,597,870,700 m, .732 of which places the three observers not just on different continents but on different planets. And it would assume they all have the same concept of "up." None of this is likely.

If we don't mind a small thing like melting the polar ice caps, let us assume for our purposes here that the Sun lies directly above the geographical North Pole. That is, an observer at the equator would not have to incline his head at all, while an observer standing at 90^o north latitude would have to tilt his head the full 90^o to see the eclipse.

The middle observer then stands somewhere at 45^o N latitude, and the far observer stands at 30^o N latitude. Let's assume their positions have the same longitude. Then the third observer is the same distance north of the middle man as the far observer is south of the middle man.

Since Latitude is a linear function of distance, it also means the near observer's latitude is as much larger than 45^o as the far observer's latitude is less than 45^o. Thus the third student stands at N latitude 60^o, and that is also the angle he must tilt his head to see the solar eclipse.

We now return the Sun to its normal location. (A place for everything and everything in its place.) We could equally well have nudged the Sun directly over the equator and played with longitudes. But the polar bears would have had a boring day.

I believe the answer is unique. If the 45^o and 30^o observers were at longitudes that differ by 180^o, so as to keep themselves and the north pole on a great circle, then an equidistant third student would be out of view of the sun. That is, for him it would be night.

[bhramarraj]

I agree to Bonanova's first post. Since the third person is equidistant from the other two and these three persons are colliniar, but they are assumed to be in the same direction to the sun, then the angle, third person will be making with the sun, will be smaller than 45 degree and greater than 30degree, if the three persons are on the same plane and collinear .

For second person say the horizontal distance from the sun is x, and vertical distance is say y. Since cot 45 degree = x/y = 1, so x = y.
Say the distance between second and third person is z, and since third person is in the middle of first and second person, so distance between first & second person will be 2z.
Then cot 30 degree 1.732 = (x + 2z)/y and since x = y, so
1 + 2z/y = 1.732 or z/y = 0.366........................[1]
Say the angle made by the third person is D degree, then cot D degree = (x+z)/y, and since x = y, so cot D = 1 + z/y........[2]
From [1] & [2], cot D = 1.366.
So D = 36.206 degree.

But the puzzles BMAD is presenting are not so easy.
The three persons might be collinear and on the same plane but their directions might be diffrent from the sun.

In that case vertical distance will remain same for each person say 'y', but the relation between the horizontal distances of the three persons from the sun is not so easy for me to calculate. I may come to some answer say tomorrow...

[BMAD]

Though Bonanova makes an excellent point on the practicality of the semi-classic geometric problem that I tried to add context to, for the sake of argument let's just assume that we are on some giant planet that has a sun in the sky at some fixed point and these three individuals are standing on a plane large enough to hold all three individuals and the sun angled just right to shine on all of them at the exact same time. I did not mean to make this problem appear so complicated but i didn't want to just give a boring find the angle in the shape problem.

I misread the order of the students.

If the light is a unit distance high, the middle observer is standing at a unit distance away.
The other two are approximately .732 units closer to and farther from the light.

cot^-1(0.268..., 1, 1.732...) = 75^o, 45^o and 30^o respectively.

However .... That is not the correct answer.

As I alluded in my first post, if we are talking about the Sun, then the unit height of the light is 1 Astronomical Unit, or 149,597,870,700 m, .732 of which places the three observers not just on different continents but on different planets. And it would assume they all have the same concept of "up." None of this is likely.

If we don't mind a small thing like melting the polar ice caps, let us assume for our purposes here that the Sun lies directly above the geographical North Pole. That is, an observer at the equator would not have to incline his head at all, while an observer standing at 90^o north latitude would have to tilt his head the full 90^o to see the eclipse.

The middle observer then stands somewhere at 45^o N latitude, and the far observer stands at 30^o N latitude. Let's assume their positions have the same longitude. Then the third observer is the same distance north of the middle man as the far observer is south of the middle man.

Since Latitude is a linear function of distance, it also means the near observer's latitude is as much larger than 45^o as the far observer's latitude is less than 45^o. Thus the third student stands at N latitude 60^o, and that is also the angle he must tilt his head to see the solar eclipse.

We now return the Sun to its normal location. (A place for everything and everything in its place.) We could equally well have nudged the Sun directly over the equator and played with longitudes. But the polar bears would have had a boring day.

I believe the answer is unique. If the 45^o and 30^o observers were at longitudes that differ by 180^o, so as to keep themselves and the north pole on a great circle, then an equidistant third student would be out of view of the sun. That is, for him it would be night.

[BMAD]

[bonanova]

If the students are looking at an eclipse of a street lamp, then they might be friends hanging out on campus, carrying sextants and measuring tape and standing close enough to discuss our answers.

If they are viewing an eclipse of the sun, then for there to be a well-defined angle of inclination, their distance from the sun must be large compared with the sun's diameter. Another way of saying this is that the sun must be more closely approximated by a point source (casting a sharp shadow) than by an extended source (casting a diffuse shadow.) This is a fair approximation at 1AU. Thus we can consider students placed at various points on the Earth.