Solving for a Side in a Triangle: Law of Cosines

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To solve for side b in a triangle using the Law of Cosines, the formula is b^2 = a^2 + c^2 - 2ac(cos B). The confusion arises because side b does not form the hypotenuse in the right triangles created by dropping a vertical line. A helpful approach is to draw an altitude to either side a or c, which can clarify the relationships between the sides and angles. Understanding the dot product and vectors may also provide additional insights into deriving the cosine rule. Visual aids, like diagrams, can significantly enhance comprehension of these geometric concepts.
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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?
 
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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?
 
DecayProduct said:
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)
 

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