MHB Triangle Side and Angle Relationship using Law of Sines

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The discussion focuses on proving the relationship between the sides and angles of a triangle using the Law of Sines. It establishes that \(\frac{A+B}{C} = \frac{\cos\left(\tfrac{1}{2}(\alpha-\beta)\right)}{\sin\left(\tfrac{1}{2}\gamma\right)}\). The proof involves manipulating the Law of Sines and applying sum-to-product identities to derive the desired equation. Key steps include expressing the sides in terms of their opposite angles and simplifying using trigonometric identities. The conclusion confirms the relationship holds true, demonstrating the interconnectedness of triangle geometry.
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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:
[sp]Law of sines[/sp]

Hint 2:
[sp]Sum to product formula[/sp]

 
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Congratulations to the following members for their correct solutions:

1) Sudharaka
2) caffeinemachine

Solution:
[sp]
The Law of Sines tells us that for any triangle $\Delta ABC$, we can relate the angles with their sides:

\[\frac{\sin\alpha}{A}= \frac{\sin\beta}{B}= \frac{\sin\gamma}{C} \]

Using this equality, we can come up with the following relations:

\[\begin{aligned}A\sin\gamma &= C\sin\alpha\\ C\sin\beta &= B\sin\gamma\end{aligned}\]

We start with the LHS and show it gives the RHS. Incorporating the relations from above, we see that

\[\begin{aligned}\frac{A+B}{C} &= \frac{A\sin\gamma + B\sin\gamma}{C\sin\gamma}\\ &= \frac{C\sin\alpha+C\sin\beta}{C\sin\gamma}\\ &= \frac{\sin\alpha + \sin\beta}{\sin\gamma}\end{aligned}\]

By the sum-to-product and double angle identities, we have

\[\begin{aligned}\frac{\sin\alpha + \sin\beta}{\sin\gamma} &= \frac{ 2\sin \left( \frac{\alpha+\beta}{2} \right) \cos \left( \frac{\alpha-\beta}{2} \right)}{\sin\gamma}\\ &= \frac{ 2\sin\left( \frac{\alpha+\beta}{2} \right) \cos\left( \frac{\alpha-\beta}{2} \right)}{2\sin \left( \frac{\gamma}{2}\right) \cos\left( \frac{\gamma}{2} \right)}\end{aligned}\]

Note that $\gamma = \pi - \alpha - \beta$. So it follows that $\cos \left(\dfrac{\gamma}{2}\right) = \cos \left( \dfrac{\pi-\alpha-\beta}{2} \right) = \sin \left( \dfrac{\alpha+\beta}{2} \right)$. Therefore, we're now left with

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

Therefore, $\dfrac{A+B}{C} = \dfrac{ \cos\left( \tfrac{1}{2}(\alpha-\beta)\right) }{\sin \left( \tfrac{1}{2}\gamma\right)}$. Q.E.D.[/sp]
 
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