Solving for a Side in a Triangle: Law of Cosines

[DecayProduct]

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 anyone 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?

 

[Cyosis]

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.

 

[DecayProduct]

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.

 

[Hogger]

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

 

[klite]

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 anyone 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)