`r_2r_3=`

Options

A. `S tan "A/2`
B. `S cot "A/2`
C. `S sin " A/2`
D. `S cos A/2`

Answer

To solve the expression \( r_2 r_3 \), where \( r_2 \) and \( r_3 \) are the radii of the excircles of a triangle, we will follow these steps:

### Step 1: Write the formulas for \( r_2 \) and \( r_3 \)

The formulas for the radii of the excircles are given by:
\[ r_2 = \frac{S}{s - b} \]
\[ r_3 = \frac{S}{s - c} \]
where \( S \) is the area of the triangle, and \( s \) is the semi-perimeter defined as:
\[ s = \frac{a + b + c}{2} \]

### Step 2: Multiply \( r_2 \) and \( r_3 \)

Now, we can find \( r_2 r_3 \):
\[ r_2 r_3 = \left( \frac{S}{s - b} \right) \left( \frac{S}{s - c} \right) = \frac{S^2}{(s - b)(s - c)} \]

### Step 3: Substitute the area \( S \)

Using Heron's formula for the area \( S \):
\[ S = \sqrt{s(s - a)(s - b)(s - c)} \]
we can substitute \( S \) into the expression for \( r_2 r_3 \):
\[ r_2 r_3 = \frac{\left( \sqrt{s(s - a)(s - b)(s - c)} \right)^2}{(s - b)(s - c)} \]

### Step 4: Simplify the expression

Now, simplifying the expression:
\[ r_2 r_3 = \frac{s(s - a)(s - b)(s - c)}{(s - b)(s - c)} \]
The \( (s - b) \) and \( (s - c) \) terms cancel out:
\[ r_2 r_3 = s(s - a) \]

### Final Result

Thus, the final result for \( r_2 r_3 \) is:
\[ r_2 r_3 = s(s - a) \]

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