Satellite angle, remote sensing

[Firben]

Homework Statement

You have a geostationary satellite (location 0◦N 0◦E, 36000 km above geoid). Assume the Earth to be spherical with a radius of REarth = 6372km

When observing a region centered at 60◦N 0◦E:

What is the incidence angle of the satellite’s line of sight (at pixel center; surface parallel≡ 0 ◦ , nadir≡ 90◦ )?

Homework Equations

φ = 60◦
r = 6370 km

The Attempt at a Solution

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http://s716.photobucket.com/user/Pitoraq/media/Rs_zpstcj5qby5.png.html?sort=3&o=0

Im not sure how to calculate the angle of incidence by that figure. This is from an exam and the solution is:

b = r × sin φ = 5517 km
a = r − r × cos φ = 3185 km
scan-angle (off nadir) β: tan β = b/(h+a) = 5517/(36000+3185) = 0.1408, β = 8.014◦
parallel surface: 90◦ − φ = 30◦
incidence-angle: γin = 90◦ − (φ + β) = 21.99◦

where do they get the relation 90◦ − φ = 30◦ from ? and where is β in my figure ?

 

[NascentOxygen]

β is the acute angle on the far right of your sketch.

 

[Firben]

But where do they get the relation γin = 90◦ − (φ + β) = 21.99◦ from ?

 

[Firben]

I got the answer by using the sinus law (v = 21.9). But what about the mean horizontal resolution in direction, respectively (in km)? You might assume the Earth to be locally flat.
If i call the right side of the figure d,then d is = 41639 km
The answer is:
tan α(lon) × d × 2 = 2.89 km
Why did they add those terms together ?

 

[NascentOxygen]

But where do they get the relation γin = 90◦ − (φ + β) = 21.99◦ from ?

From there being 180° in a ∆. Your diagram is too small; you need a large, neatly drawn figure that is so spacious that you can imagine yourself walking around in it----that's how I view geometry.
γ_in needs to be shown measured against a tangent to the Earth's surface, so there's your 90° between a radius and that tangent.

 

[Firben]

Yes, i saw it when i plotted a new figure. But where do the relation tan α(lon) × d × 2 = 2.89 km came from ?