solut2 - Solutions to homework # 2 I.15. The area of a...

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Unformatted text preview: Solutions to homework # 2 I.15. The area of a triangle is equal to half the area of the parallelogram defined by two sides of the triangle and the angle between them. The latter area is the absolute value of the cross product of the two vectors corresponding to the two sides. Taking all three pairs of the sides and comparing the obtained area, we get 1 2 AB sin = 1 2 AC sin = 1 2 BC sin = Area( ABC ) . Dividing through by 1 2 ABC , we obtain sin A = sin B = sin C . I.16. Let a , b and c denote the sides of a triangle and let ~ a , ~ b , and ~ c denote the corresponding vectors directed so that ~ a = ~ b + ~ c . Then c 2 = ~ c ~ c = ( ~ a- ~ b ) ( ~ a- ~ b ) = ~ a ~ a- 2 ~ a ~ b + ~ b ~ b = a 2- 2 ab cos + b 2 where is the angle between ~ a and ~ b . I.25a. By Stokes (Greenes) theorem, I C ~ u d ~ l = Z Z S ~ ~ u dA where the second integral is over the area enclosed by the curve C . For this problem, ~ ~ u = (- xy 2 ) x...
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This note was uploaded on 12/12/2009 for the course MATHEWS SOC 2000 taught by Professor M during the Spring '09 term at AIB College of Business.

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solut2 - Solutions to homework # 2 I.15. The area of a...

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