How to prove the law of cosines for all types of triangles?

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To prove the law of cosines, the equation a^2 = b^2 + c^2 - 2bc(cosA) can be verified by dropping a perpendicular from angle B to side b, creating two right triangles. Trigonometry can be used to find the length of this perpendicular, which relates to the sides of the triangles formed. Applying the Pythagorean theorem to these right triangles will yield useful relationships between the sides. This method is applicable for all types of triangles, including obtuse, acute, and right triangles. Ultimately, the goal is to demonstrate the truth of the equation through these geometric principles.
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Homework Statement


a^2 = b^2 + c^2 - 2bc(cosA)

Confirm that this is true.

Hint: Drop a perpendicular from angle B to side b and use the two right triangles formed.

Homework Equations





The Attempt at a Solution



Honestly, I am looking at my diagram and simply cannot figure out what to do.
 
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Can you use trigonometry to determine the length of the perpendicular that it's suggested that you draw? Is there a formula that would let you relate this to the other sides of the right triangles formed?
 
I'm pretty sure that you can use trigonometry to determine the length of the perpendicular. But I only know that you have to show that this equation is true.
 
vortex193 said:
I'm pretty sure that you can use trigonometry to determine the length of the perpendicular. But I only know that you have to show that this equation is true.

The equation that you're trying to verify relates the squares of the length of the sides of the original triangle to each other. Are any of these sides shared with the right triangles that the hint tells you to form? Is there a chance that applying the Pythagorean theorem to these right triangles would provide useful information for this problem?
 
Try drawing in the height of the triangle, then utilize the Pythagorean theorem. You should be able to prove this for obtuse, acute and right triangles.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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