Sunday, May 15, 2011

Law of sines, and cosines



Continuing with some homework, we're going to use vector algebra to prove two geometric theorems.

[This is a similar diagram to the one from last time, but with labels switched around--sorry for any confusion].

We have vectors (a, b and c) and the lengths of the corresponding sides (|a| = a, etc.); also, the angle opposite side a is labeled A and so on.

The law of sines states that the ratio of the length of each side to the sine of the angle opposite is the same:

a/sin(A) = b/sin(B) = c/sin(C)

Recall that the area of the parallelogram formed by a and b is given by the absolute value of the cross-product:

|a X b| = a b sin(C)

And the area of the triangle is one-half that. But we must obtain the same area no matter which two vectors we use to compute the cross-product, and no matter which orientation. Thus:

|c X -a| = a c sin(B)
|-b X c| = b c sin(A)


also have the same area.

a b sin(C) = a c sin(B) = b c sin(A)

This leads directly to the law of sines. Now let's relabel the triangle slightly.


If b - a looks a little funny, just consider that

a + b - a = b

The law of cosines states that:

c2 = a2 + b2 - 2 a b cos(C)

where c = |b - a|.
We can obtain this simply by expanding the dot product:

(b - a) • (b - a) =
= bb - ba - ab + aa
= b2 + a2 - 2 a b cos(C)


But

(b - a) • (b - a) = c2

So finally:

c2 = b2 + a2 - 2 a b cos(C)

Using vectors makes it easy.

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