ABC is an equilateral triangle inscribed in a circle with AB = 5 cm. Let the bisector of angle A meets BC in X and the circle in Y. What is the value of $AX\times AY$?
(A) 16 sq. cm
(B) 20 sq. cm
(C) 25 sq. cm
(D) 30 sq. cm

Hint: We solve this question by considering the triangle ACX and apply the trigonometric ratio $\cos \theta =\dfrac{\text{Adjacent Side}}{\text{Hypotenuse}}$ and substitute the angle CAX and side AC to find the value of length of AX. Then we consider the triangle ACY and apply the trigonometric ratio cosine to the angle CAY using the formula $\cos \theta =\dfrac{\text{Adjacent Side}}{\text{Hypotenuse}}$ and substitute the values of angle CAY and AC to find the length of AY. Then we multiply the values of AX and AY to find the required answer.

Complete step-by-step solution:
Given that the triangle ABC is equilateral with side AB = 5 cm and is inscribed in a circle.
Given that bisector of an angle, A intersects side BC at X. So, we have,
$\angle CAX={{30}^{\circ }}$
As ABC is an equilateral triangle angular bisector is also a median and also a perpendicular bisector. So, we get
$\angle AXC={{90}^{\circ }}$
So, triangle ACX is right-angled.
Now let us plot a rough diagram from the given information.

In triangle ACY, the side AY passes through the center of the circle. So, the triangle ACY is right-angled at C, which is $\angle ACY={{90}^{\circ }}$.
Now, let us consider the right-angled triangle ACX.
Let us consider the formula for cosine, $\cos \theta =\dfrac{\text{Adjacent Side}}{\text{Hypotenuse}}$
So, applying the above formula we get,
$\Rightarrow \cos \left( \angle CAX \right)=\dfrac{AX}{AC}$
As ABC is an equilateral triangle AB=BC=CA=5 cm. So, substituting them in the above we get,
$\begin{align}
  & \Rightarrow \cos {{30}^{\circ }}=\dfrac{AX}{5} \\
 & \Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{AX}{5} \\
 & \Rightarrow AX=\dfrac{5\sqrt{3}}{2}cms...........\left( 1 \right) \\
\end{align}$

Now, let us consider the right-angled triangle ACY.
Let us consider the formula for cosine, $\cos \theta =\dfrac{\text{Adjacent Side}}{\text{Hypotenuse}}$
So, applying the above formula we get,
$\Rightarrow \cos \left( \angle CAY \right)=\dfrac{AC}{AY}$
As $\angle CAY=\angle CAX$, we get $\angle CAY={{30}^{\circ }}$.
Substituting this value and the value of AC in the above equation we get,
$\begin{align}
  & \Rightarrow \cos {{30}^{\circ }}=\dfrac{5}{AY} \\
 & \Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{5}{AY} \\
 & \Rightarrow AY=\dfrac{10}{\sqrt{3}}cms...........\left( 2 \right) \\
\end{align}$
As we need to find the value of $AX\times AY$, let us substitute the values of AX and AY obtained in the equations (1) and (2). Then we get,
$\begin{align}
  & \Rightarrow AX\times AY=\dfrac{5\sqrt{3}}{2}\times \dfrac{10}{\sqrt{3}} \\
 & \Rightarrow AX\times AY=5\times 5 \\
 & \Rightarrow AX\times AY=25sq.cms \\
\end{align}$
So, we get the value of $AX\times AY$ as 25 sq. cm
Hence answer is Option C.

Note: There is a possibility of making a mistake by taking the formula for the trigonometric ratio as $\sin \theta =\dfrac{\text{Adjacent Side}}{\text{Hypotenuse}}$. But the actual formula for sine is $\sin \theta =\dfrac{\text{Opposite}}{\text{Hypotenuse}}$ and for cosine is $\cos \theta =\dfrac{\text{Adjacent Side}}{\text{Hypotenuse}}$.