Question:
In a triangle ABC, if \(a = 13, b = 14, c = 15\), then
\[
\sin \frac{A}{2} = ?
\]
In a triangle ABC, if \(a = 13, b = 14, c = 15\), then
\[
\sin \frac{A}{2} = ?
\]
Show Hint
For half-angle formulas in triangles, compute \(s\) and \(s-a, s-b, s-c\) first, then apply \(\sin(A/2) = \sqrt{\frac{(s-b)(s-c)}{bc}}\).
Updated On: Jul 18, 2026
- \(\frac{1}{\sqrt{5}}\)
- \(\frac{1}{\sqrt{7}}\)
- \(\frac{3}{4}\)
- \(\frac{4}{5}\)
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The Correct Option is A
Solution and Explanation
Step 1: Compute semi-perimeter.
\[ s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = 21 \]
Step 2: Compute \(s-a, s-b, s-c\).
\[ s-a = 21-13=8, \quad s-b = 21-14=7, \quad s-c = 21-15=6 \]
Step 3: Use half-angle formula.
\[ \sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}} = \sqrt{\frac{7 \cdot 6}{14 \cdot 15}} \]
Step 4: Simplify fraction.
\[ \sqrt{\frac{42}{210}} = \sqrt{\frac{1}{5}} \]
Step 5: Final conclusion.
\[ \sin \frac{A}{2} = \frac{1}{\sqrt{5}} \]
\[ \boxed{\frac{1}{\sqrt{5}}} \]
\[ s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = 21 \]
Step 2: Compute \(s-a, s-b, s-c\).
\[ s-a = 21-13=8, \quad s-b = 21-14=7, \quad s-c = 21-15=6 \]
Step 3: Use half-angle formula.
\[ \sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}} = \sqrt{\frac{7 \cdot 6}{14 \cdot 15}} \]
Step 4: Simplify fraction.
\[ \sqrt{\frac{42}{210}} = \sqrt{\frac{1}{5}} \]
Step 5: Final conclusion.
\[ \sin \frac{A}{2} = \frac{1}{\sqrt{5}} \]
\[ \boxed{\frac{1}{\sqrt{5}}} \]
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