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

  • Thread starter Thread starter maxkor
  • Start date Start date
  • Tags Tags
    Angles Triangle
AI Thread Summary
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.
maxkor
Messages
79
Reaction score
0
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?
 
Mathematics news on Phys.org
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)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top