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:

`c`^{2} = a^{2} + b^{2} - 2 a b cos(C)

where

`c = |`**b** - **a**|

.We can obtain this simply by expanding the dot product:

`(`**b** - **a**) • (**b** - **a**) =

= **b** • **b** - **b** • **a** - **a** • **b** + **a** • **a**

= b^{2} + a^{2} - 2 a b cos(C)

But

`(`**b** - **a**) • (**b** - **a**) = c^{2}

So finally:

`c`^{2} = b^{2} + a^{2} - 2 a b cos(C)

Using vectors makes it easy.

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