Question:

Step 1: Configuration of the circles. Let $C_1$ be the circle passing through vertex $A$ and touching side $BC$ at point $B$. Let its radius be $r_1$. Let $C_2$ be the circle passing through vertex $A$ and touching side $BC$ at point $C$. Let its radius be $r_2$. Step 2: Determine the radius $r_1$. Let $O_1$ be the center of $C_1$. Since $C_1$ is tangent to $BC$ at $B$, the radius $O_1B$ is perpendicular to $BC$. Place $B$ at the origin $(0,0)$ and let $BC$ lie along the positive x-axis. Then $O_1$ has coordinates $(0, r_1)$. Let $A$ have coordinates $(x_A, y_A)$. The distance $AB = \sqrt{x_A^2 + y_A^2}$. In $\triangle ABC$, the coordinates of $A$ can be expressed as $(AB \cos B, AB \sin B)$, where $B = \angle ABC$. Since $A$ lies on $C_1$, the distance $O_1A$ must be equal to $r_1$. $$ (AB \cos B - 0)^2 + (AB \sin B - r_1)^2 = r_1^2 $$ $$ AB^2 \cos^2 B + AB^2 \sin^2 B - 2 AB \sin B r_1 + r_1^2 = r_1^2 $$ $$ AB^2 - 2 AB \sin B r_1 = 0 $$ Since $AB \neq 0$, we can divide by $AB$: $$ AB = 2 \sin B r_1 $$ Therefore, $$ r_1 = \frac{AB}{2 \sin B} $$ Step 3: Determine the radius $r_2$. Similarly, let $O_2$ be the center of $C_2$. Since $C_2$ is tangent to $BC$ at $C$, the radius $O_2C$ is perpendicular to $BC$. Place $C$ at the origin $(0,0)$ and let $BC$ lie along the negative x-axis. Then $O_2$ has coordinates $(0, r_2)$. Let $A$ have coordinates $(x_A', y_A')$. The distance $AC = \sqrt{(x_A')^2 + (y_A')^2}$. In $\triangle ABC$, the coordinates of $A$ can be expressed as $(AC \cos C, AC \sin C)$, where $C = \angle ACB$. Since $A$ lies on $C_2$, the distance $O_2A$ must be equal to $r_2$. $$ (AC \cos C - 0)^2 + (AC \sin C - r_2)^2 = r_2^2 $$ $$ AC^2 \cos^2 C + AC^2 \sin^2 C - 2 AC \sin C r_2 + r_2^2 = r_2^2 $$ $$ AC^2 - 2 AC \sin C r_2 = 0 $$ Since $AC \neq 0$, we can divide by $AC$: $$ AC = 2 \sin C r_2 $$ Therefore, $$ r_2 = \frac{AC}{2 \sin C} $$ Step 4: Calculate the product $r_1 r_2$. Multiplying the expressions for $r_1$ and $r_2$: $$ r_1 r_2 = \left(\frac{AB}{2 \sin B}\right) \left(\frac{AC}{2 \sin C}\right) = \frac{AB \cdot AC}{4 \sin B \sin C} $$ By the Law of Sines in $\triangle ABC$, we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$, where $a=BC$, $b=AC$, and $c=AB$. From this, we can express $\sin B$ and $\sin C$ in terms of $a$, $b$, $c$, and $\sin A$: $$ \sin B = \frac{b \sin A}{a} \quad \text{and} \quad \sin C = \frac{c \sin A}{a} $$ Substitute these into the expression for $r_1 r_2$: $$ r_1 r_2 = \frac{c \cdot b}{4 \left(\frac{b \sin A}{a}\right) \left(\frac{c \sin A}{a}\right)} = \frac{bc}{4 \frac{bc \sin^2 A}{a^2}} = \frac{a^2}{4 \sin^2 A} $$ Step 5: Calculate $\sqrt{r_1 r_2}$ using the given values. We are given $a = 5$ and $\angle A = 30^\circ$. $$ \sqrt{r_1 r_2} = \sqrt{\frac{a^2}{4 \sin^2 A}} = \frac{a}{2 \sin A} $$ Substitute the given values: $$ \sqrt{r_1 r_2} = \frac{5}{2 \sin 30^\circ} = \frac{5}{2 \cdot (1/2)} = \frac{5}{1} = 5 $$