MHB Solving Acute Triangle Angles Given $A,B,C$ & Equations

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To solve for \( \cos(A+C) \) in the acute triangle with angles \( A, B, C \), the equations \( (5+4\cos A)(5-4\cos B)=9 \) and \( (13-12\cos B)(13-12\cos C)=25 \) are given. It is noted that \( A+B+C=180^{\circ} \) and \( \cos B=-\cos(A+C) \). A recommended approach is to isolate \( \cos B \) from both equations to facilitate further calculations. This method has proven effective for others in similar problems. The discussion emphasizes the importance of manipulating the equations to find the desired cosine value.
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Given that $A,B,C$ be angles in an acute triangle.

If $(5+4\cos A)(5-4\cos B)=9$ and $(13-12\cos B)(13-12\cos C)=25$

find $cos(A+C)$.
I know $A+B+C=180^{o}$ and $\cos B=-\cos(A+C)$ and what next?
 
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Hi, maxkor!(Wave)

Thankyou for sharing your problem on the MHB site!

You ask for the next step. One way (which worked for me) would be to isolate $\cos B$ in both equations.

I suggest, you try this out(Nod)
 
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