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 \(\sin A:\sin B:\sin C\) equal to?
\(7:6:5\)
Setting each equal ratio to \(k\): \(a+b=13k\), \(b+c=11k\), \(c+a=12k\). Adding all three, \(a+b+c=18k\), so \(a=7k\), \(b=6k\), \(c=5k\). By the sine rule, \(\sin A:\sin B:\sin C=a:b:c=7:6:5\).
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