Question

In \(△ABC\), with usual notation, if \(a = 13,b = 14,c = 15\), then the sum of the values of \(sin(\frac{A}{2})\) and \(sinA\) is....

• A. \(\frac{14}{5\sqrt{5}}\)
• B. \(\frac{\sqrt{5}+4}{5}\)
• C. \(\frac{5+\sqrt{5}}{5}\)
• D. \(\frac{2}{\sqrt{5}}+\frac{1}{2}\)

Hint:

Use the cosine rule for cos A, then half angle formula.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Angle \(A\) is opposite side \(a = 13\). Use \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\).

Step 2: Key Formula or Approach:
\(\sin A = \sqrt{1 - \cos^2 A}\) and \(\sin\frac{A}{2} = \sqrt{\frac{1 - \cos A}{2}}\).

Step 3: Detailed Explanation:
\(\cos A = \frac{196 + 225 - 169}{2 \times 14 \times 15} = \frac{252}{420} = \frac35\).
\(\sin A = \sqrt{1 - \frac{9}{25}} = \frac45\).
\(\sin\frac{A}{2} = \sqrt{\frac{1 - \frac35}{2}} = \sqrt{\frac15} = \frac{1}{\sqrt5}\) (positive since \(\frac A2\) is acute).
\[ \sin\frac{A}{2} + \sin A = \frac{1}{\sqrt5} + \frac45 = \frac{\sqrt5}{5} + \frac45 = \frac{\sqrt5 + 4}{5} \]

Final Answer:
The sum is \(\frac{\sqrt{5} + 4}{5}\), option (B). \[ \boxed{\frac{\sqrt{5}+4}{5}} \]