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: For a triangle with semiperimeter \(s\), area \(\Delta\), and side \(c\), \[ \Delta=rs, \] where \(r\) is the inradius. Also, \[ \sin\frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{ab}}. \] A more useful relation is \[ \tan\frac{C}{2} = \frac{r}{s-c}. \]

Step 1: Find the inradius \(r\). Using \[ \Delta=rs, \] we get \[ 10\sqrt2=r(10). \] Hence, \[ r=\sqrt2. \]

Step 2: Find \(\tan\frac{C}{2}\). Given \[ c=9, \qquad s=10. \] Therefore, \[ s-c=10-9=1. \] Using \[ \tan\frac{C}{2} = \frac{r}{s-c}, \] we obtain \[ \tan\frac{C}{2} = \frac{\sqrt2}{1} = \sqrt2. \]

Step 3: Find \(\sin\frac{C}{2}\). Using \[ \sin\theta = \frac{\tan\theta} {\sqrt{1+\tan^2\theta}}, \] with \[ \theta=\frac{C}{2}, \] we get \[ \sin\frac{C}{2} = \frac{\sqrt2} {\sqrt{1+2}}. \] \[ = \frac{\sqrt2}{\sqrt3}. \] \[ = \sqrt{\frac23}. \]

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