How Do You Calculate the Altitude and Distance of Spacecraft Using Trigonometry?

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I know this is simple, but I need help. I've attached the image (of ISS and SS with angles, etc..)


Based upon the information provided on International Space Station (I.S.S.) and Space Shuttle (S.S.), and tracking stations C and D, answer following questions:

a) What are the altitudes of the space station and space shuttle?
b) How far apart are they? (horizontal and vertical distances)
c) How far is each space vehicle from each of the tracking stations?
 
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You need to show us some of your work in order for us to help you (PF homework forum rules). How would you go about starting these problems?
 
I did it this far, but I'm not sure if I even did it right:

Solution:

Laws used: Sin = Opposite / Hypotenuse, Cos = adjacent / hypotenuse and A / sina = B/ sin b = C / sin c

< A = 180 - 33 - 72
= 75

< B = 180 - 31 - 81
= 68

d --> d/sin72
= 1418/sinA
= 1418/sinA*sin72
= 1418/sin75*sin72
= 1418/0.9659*0.95105
= 1418/0.918619
= 1543.62
c --> c/sin31
= 1418/sinB
= 1418/sin31*sin68
= 1418/0.51503*0.92718
= 1418/0.4775255
= 2969.475

a1 (altitude 1) => sin33 = a1/d
a1 = sin33*d
= 0.544639*1543.62
= 840.7
a2 (altitude 2) => sin81 = a2/c
a2 = sin81*c
= 0.9877*2969.475
= 2932.5
f --> cos33 = f/d
f = cos33*d
= 0.8387*1543.54
= 1294.6
g --> cos81 = g/c
g = cos81*c
= 0.15643*2969.475
= 464.44

b) H (horizontal distance) = 1418 - f - g
= 1418 - 1294.6 - 464.52
= -341.12
V (vertical distance) = a2 - a1
= 2932.5 - 840.715
= 2091.785
 
c) Distance from ISS to Houston = d => 1543.62
Distance from SS to Houston = c => c/sin33
= 1418/sinB
= 1418/sinB*sin33
= 1418/sin68*sin33
= 1418/0.92718*0.544639
= 1418/0.504978
= 2808.04
 
I think you made a mistake when you rearranged your equations using the sine law to find d and c, giving you the wrong values to work with for the rest of the question (ex. d=(1418*sin72)/sin75). You repeat this mistake in part (c), as well as mixing up some angles I think.