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- #1

- Feb 14, 2012

- 3,838

- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,838

- Thread starter
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- #3

- Feb 14, 2012

- 3,838

Yes,Am I correct in assuming that you're using a, b, c to represent the lengths of the triangle and A, B, C to represent the angles opposite their corresponding letter side?

- May 31, 2013

- 118

Sorry I don't know how to use $\LaTeX$...

We get:

\(\displaystyle 2\sin(B)=\sin(A)+\sin(C)\)

Hence:

\(\displaystyle 2\left(2\sin\left(\frac{B}{2} \right)\cos\left(\frac{B}{2} \right) \right)=2\sin\left(\frac{A+C}{2} \right)\cos\left(\frac{A-C}{2} \right)

\)

By solving:

\(\displaystyle 2\cos\left(\frac{B}{2} \right)=\cos\left(\frac{A-C}{2} \right)\)

And we get:

\(\displaystyle \cos(A-C)=3-4\cos(B)\)

And the final result is 3...am I correct?

We get:

\(\displaystyle 2\sin(B)=\sin(A)+\sin(C)\)

Hence:

\(\displaystyle 2\left(2\sin\left(\frac{B}{2} \right)\cos\left(\frac{B}{2} \right) \right)=2\sin\left(\frac{A+C}{2} \right)\cos\left(\frac{A-C}{2} \right)

\)

By solving:

\(\displaystyle 2\cos\left(\frac{B}{2} \right)=\cos\left(\frac{A-C}{2} \right)\)

And we get:

\(\displaystyle \cos(A-C)=3-4\cos(B)\)

And the final result is 3...am I correct?

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- Feb 14, 2012

- 3,838

You're not only correct...you're brilliant!Sorry I don't know how to use $\LaTeX$...

We get:

\(\displaystyle 2\sin(B)=\sin(A)+\sin(C)\)

Hence:

\(\displaystyle 2\left(2\sin\left(\frac{B}{2} \right)\cos\left(\frac{B}{2} \right) \right)=2\sin\left(\frac{A+C}{2} \right)\cos\left(\frac{A-C}{2} \right)

\)

By solving:

\(\displaystyle 2\cos\left(\frac{B}{2} \right)=\cos\left(\frac{A-C}{2} \right)\)

And we get:

\(\displaystyle \cos(A-C)=3-4\cos(B)\)

And the final result is 3...am I correct?