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homework1solutions

# homework1solutions - AST 3722C Spring 2008 Homework#1...

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AST 3722C - Spring 2008 Homework #1 – Solutions 1. [1 point] Chapter 7 Problem 5, pages 57-58. Solve completely the spherical triangle ABC , given: (i) a = 37 48 , b = 29 51 , C = 74 37 . In the standard notation, a and b refer to sides, and C refers to the enclosed angle. Thus the problem asks you to find A , B , and c , the other two angles and the third side. Often it helps to draw the triangle to help visualize what’s going on; check out the crudely drawn triangles in Figure 1. In any case the third side c can be derived from the law of cosines: cos c = cos a cos b + sin a sin b cos C = 0 . 7663 c = 39 . 98 = 39 59 . Then the other two sides can be derived from the law of sines: sin A sin a = sin C sin c sin A = sin a sin C sin c = 0 . 9197 A = 66 . 88 = 66 53 . And finally: sin B sin b = sin C sin c sin B = sin c sin C sin c = 0 . 7469 B = 48 . 32 = 48 19 . (ii) b = 98 23 , c = 76 39 , A = 52 23 . They’ve permuted the given info but it is essentially the same – two sides and an included angle. So the procedure is the same as for part (i), you just need to start with a permuted law of cosines: cos a = cos b cos c + sin b sin c cos A . In this case you should find that 1

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Figure 1 – Roughly approximately appropriate spherical triangles for Problem 1. cos a = 0 . 5539 , a = 56 . 37 = 56 22 , sin B = 0 . 9412 , B = 70 . 25 = 70 15 , sin C = 0 . 9257 , C = 67 . 77 = 67 46 . (iii) c = 30 57 , a = 127 08 , B = 138 19 . Same business, everything’s just permuted again: cos b = 0 . 8239 , b = 145 . 48 = 145 29 , sin C = 06035 , C = 37 . 12 = 37 07 , sin A = 0 . 9355 , A = 69 . 32 = 69 19 . (iv) a = 90 , b = 74 39 , C = 165 29 . Same business, everything’s just permuted again: cos c = 0 . 9335 , c = 158 . 99 = 159 00 , sin A = 0 . 6993 , A = 44 . 37 = 44 22 , sin B = 0 . 6743 , B = 42 . 40 = 42 24 . Do note that these answers are given on page 460 of your textbook. 2
2. [1 point] Chapter 7 Problem 9, page 58. Note here (as I mentioned on the webpage) that the longitudes really should be negative. These two locations are west of Greenwich and so in the convention we’ve been using in class (and that is common in astronomy), the longitudes should be 2 18 . 4 for Jodrell Bank and 79 50 . 2 for Green Bank. Check out Figure 2 for a representation of the globe showing where the two Banks are with respect to the Prime Meridian. Figure 2 – Location of Green Bank and Jodrell Bank with respect to the Prime Meridian; notice they are both west of it, not east. Generally speaking, spherical astronomy problems can be solved most easily if you setup a spherical triangle where the third vertex is a special point. Often that point will be a pole of the coordinate system you’re in. In this case we’re on Earth so the third point should be the North Pole. It could be the South Pole too if you want but that would make things slightly harder. Some of you had the idea to let the third vertex be south of Jodrell Bank and east of Green Bank such that you would make a right spherical triangle. The right angle would be the angle opposite the arc connecting to the two Banks. This method can work,

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homework1solutions - AST 3722C Spring 2008 Homework#1...

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