Question

In \(\triangle ABC\), if \[ c=9,\qquad s=10,\qquad \Delta=10\sqrt2, \] then \[ \sin\frac{C}{2} = \]

• A. \(\dfrac12\)
• B. \(\sqrt{\dfrac23}\)
• C. \(\dfrac{\sqrt3-1}{2\sqrt2}\)
• D. \(\dfrac13\) \bigskip

Hint:

When \(s\), \(\Delta\), and a side are given, first find the inradius using \[ r=\frac{\Delta}{s}. \] Then use \[ \tan\frac{A}{2}=\frac{r}{s-a}, \] which often leads directly to the required half-angle value.

Answer 1

The Correct Option is B

Concept: Use the formula relating inradius \(r\), semiperimeter \(s\), and area \(\Delta\): \(r = \frac{\Delta}{s}\). Then use the half-angle formula \(\sin\frac{C}{2} = \frac{r}{\sqrt{r^2 + (s-c)^2}}\).

Step 1: Find inradius \(r\). Given \(\Delta = 10\sqrt{2}, s = 10\). \[ r = \frac{\Delta}{s} = \frac{10\sqrt{2}}{10} = \sqrt{2}. \]

Step 2: Compute \(s-c\). Given \(c = 9\), so \(s - c = 10 - 9 = 1\).

Step 3: Find \(\sin\frac{C}{2}\). From right triangle geometry, \(\tan\frac{C}{2} = \frac{r}{s-c} = \frac{\sqrt{2}}{1} = \sqrt{2}\). Alternatively, directly: \[ \sin\frac{C}{2} = \frac{r}{\sqrt{r^2 + (s-c)^2}} = \frac{\sqrt{2}}{\sqrt{2 + 1}} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}. \]

Step 4: Write the final answer. \[ \boxed{\sqrt{\frac23}} \]