Solving for a Side in a Triangle: Law of Cosines

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Homework Help Overview

The discussion revolves around the application of the Law of Cosines in triangle geometry, specifically focusing on solving for a side of a triangle given its angles and other sides. The original poster expresses confusion about deriving the formula for side b and angle B, particularly since side b does not serve as the hypotenuse in the context of their approach.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the derivation of the Law of Cosines and the challenges faced when attempting to apply it to side b. The original poster describes their method of dividing side b and deriving formulas for sides a and c but seeks guidance on how to approach side b. Others suggest drawing an altitude or providing a visual representation to clarify the situation.

Discussion Status

The conversation is ongoing, with participants exploring different methods and seeking clarification on the original poster's approach. Suggestions for visual aids and alternative methods indicate a productive exchange, though no consensus has been reached on the best way to derive the formula for side b.

Contextual Notes

The original poster notes that their question is more general and not strictly a homework problem, which may influence the nature of the discussion and the responses provided.

DecayProduct
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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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