 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2/1/a/solution

Choose a common origin at vertex

C

a = CB

and

b = C A

, and orient the remaining edge as

c = A B

. Then

c = a - b .

This convention makes the interior angle between

a

and

b

equal to

C

, and their lengths equal to the opposite side lengths

a

and

b

Take the dot product of the difference with itself. Since

a \cdot b = ab cos C

, the law of cosines follows:

c^{2} = a^{2} + b^{2} - 2 ab cos C .

The magnitude of the cross product of two edge vectors is twice the triangle's area

K

. Applying this at each vertex gives

2 K = ab sin C = b c sin A = c a sin B

. Divide by the nonzero product

ab c

to obtain the law of sines:

\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c} .

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