For the next two (02) items that follow : In a triangle \(ABC\), \(\dfrac{a+b}{13}=\dfrac{b+c}{11}=\dfrac{c+a}{12}\).
What is \(\cos A:\cos B:\cos C\) equal to?
\(7:19:25\)
Taking \(a=7,b=6,c=5\), the cosine rule gives \(\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{1}{5}\), \(\cos B=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{19}{35}\), and \(\cos C=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{5}{7}\). Writing these over a common denominator of 35 gives \(\cos A:\cos B:\cos C=7:19:25\).
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