Question

In a triangle ABC, $a=5, b=4$ and $\tan\frac{C}{2} = \sqrt{\frac{7}{9}}$, then its inradius r =

• A. $\frac{\sqrt{7}}{2}$
• B. $2\sqrt{7}$
• C. $\frac{9}{\sqrt{7}}$
• D. $\frac{4}{\sqrt{7}}$

Hint:

For problems involving sides and half-angles of a triangle, the half-angle formulas for sine, cosine, and tangent are indispensable. The formula $r = (s-c)\tan(C/2)$ provides a direct link between the inradius and the half-angle.

Answer 1

The Correct Option is A

Step 1: Use the inradius formula.
\[ r = (s-c) \tan\frac{C}{2}, s = \frac{a+b+c}{2} \]
Step 2: Apply the half-angle formula.
\[ \tan^2\frac{C}{2} = \frac{(s-a)(s-b)}{s(s-c)} = \frac{7}{9} \]
Step 3: Solve for $c$.
\[ s = \frac{9+c}{2}, s-a = \frac{c-1}{2}, s-b = \frac{c+1}{2}, s-c = \frac{9-c}{2} \] \[ \frac{7}{9} = \frac{(c-1)(c+1)/4}{(9+c)(9-c)/4} = \frac{c^2-1}{81-c^2} \implies 2x^2 + 3x - 1 = 0 \] \[ 7(81-c^2) = 9(c^2-1) \implies c^2 = 36 \implies c = 6 \]
Step 4: Compute $r$.
\[ s = \frac{15}{2}, s-c = \frac{3}{2}, r = \frac{3}{2} \cdot \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{2} \]