Question

In \(\Delta ABC\), with the usual notations, if \(\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{6}=1\), then \(\frac{\sin A}{\sin B}=\ldots\)

• A. \(\frac{7}{8}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{5}{9}\)
• D. \(\frac{8}{7}\)

Hint:

Solve for the side lengths from the three given sums, then use the sine rule a/b = sin A / sin B.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The given conditions are \(a+b=7\), \(b+c=8\) and \(c+a=6\).

Step 2: Find the sides
Adding: \(2(a+b+c)=21\), so \(a+b+c=\frac{21}{2}\).
\(c=\frac{21}{2}-7=\frac{7}{2}\), \(a=\frac{21}{2}-8=\frac{5}{2}\), \(b=\frac{21}{2}-6=\frac{9}{2}\).

Step 3: Apply the sine rule
\(\dfrac{\sin A}{\sin B}=\dfrac{a}{b}=\dfrac{5/2}{9/2}=\dfrac{5}{9}\)

Step 4: Compare the options
The value 5/9 is the third option.

Final Answer:
The ratio sin A / sin B equals 5/9. \[ \boxed{\text{(C) }\dfrac{5}{9}} \]