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

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Unformatted text preview: 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 whats 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 . Theyve 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 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, everythings just permuted again: cos b = . 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, everythings just permuted again: cos c = . 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 weve 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....
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This note was uploaded on 11/09/2009 for the course AST 4700 taught by Professor Fernandez during the Spring '09 term at University of Central Florida.

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

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