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- Jan 26, 2012

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Thank you to Chris L T521 for submitting this problem.

Let $A$, $B$ and $C$ be sides of a triangle, and let $\alpha$ be the angle opposite of side $A$, $\beta$ be the angle opposite of side $B$ and $\gamma$ be the angle opposite of side $C$. Show that

\[\frac{A+B}{C} = \frac{\cos\left(\tfrac{1}{2}(\alpha-\beta)\right)}{\sin\left(\tfrac{1}{2}\gamma\right)}\]

Hint:

Hint 2:

Remember to read the POTW submission guidelines to find out how to submit your answers!

Let $A$, $B$ and $C$ be sides of a triangle, and let $\alpha$ be the angle opposite of side $A$, $\beta$ be the angle opposite of side $B$ and $\gamma$ be the angle opposite of side $C$. Show that

\[\frac{A+B}{C} = \frac{\cos\left(\tfrac{1}{2}(\alpha-\beta)\right)}{\sin\left(\tfrac{1}{2}\gamma\right)}\]

Hint:

Law of sines

Hint 2:

Sum to product formula

Remember to read the POTW submission guidelines to find out how to submit your answers!

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