# Cosine Law

This is more of a general question, than it is a homework problem.

If I have a triangle, angles A, B, and C, and corresponding sides a, b, and c, and I want to solve for any one side I understand we use the law of cosines. So far I have been able to derive two of the formulae by dropping a vertical line to divide side b into two parts, x, and b-x. Sides a and c form the hypotenuses of the two right triangles formed by dividing b.

Doing a little algebraic magic, I get:

a^2 = b^2 + c^2 - 2bc cos A and
c^2 = a^2 + b^2 - 2ab cos C

I am hung up on how to do this for side b and angle B. Side b does not form the hypotenuse of right triangle, and so I'm confused as to how to go about this. Anyone have a pointer?

## Answers and Replies

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Cyosis
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Adding a picture of the type of triangle you've drawn would help a great deal. On a side note do you know what a dot product is and how to add up vectors, because if you do there is a very easy way to derive the cosine rule.

OK, I whipped up a picture in paint. Sorry for its crudeness. As far as dot products and vectors, well, I haven't gotten there yet. I know of them is the most cursory way. I know what a vector is, but I don't know the math.

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can't you just draw an altitude to either side a or c?

This is more of a general question, than it is a homework problem.

If I have a triangle, angles A, B, and C, and corresponding sides a, b, and c, and I want to solve for any one side I understand we use the law of cosines. So far I have been able to derive two of the formulae by dropping a vertical line to divide side b into two parts, x, and b-x. Sides a and c form the hypotenuses of the two right triangles formed by dividing b.

Doing a little algebraic magic, I get:

a^2 = b^2 + c^2 - 2bc cos A and
c^2 = a^2 + b^2 - 2ab cos C

I am hung up on how to do this for side b and angle B. Side b does not form the hypotenuse of right triangle, and so I'm confused as to how to go about this. Anyone have a pointer?
$$b^2 = a^2 + c^2 - 2ac(cos B)$$