Question:
Regular hexagon inscribed in a unit circle
(A0A1).(A0A2).(A0A4) = ?

a)" "1

b)" "sqrt(3)

c)" "3

d)" "3Sqrt(3)

Solution:
Since this is a regular hexagon so all the interior triangles will be equilateral.
Also In Triangle (OA0A4) ,
(A0A1) = 1 (since triangle is equilateral and r = 1)

Similarly ,
(A0A2) = 1 = .......(A0A5)

Applying cosine formula in triangle
(A0A1A2) we get :
(A0A2) = sqrt(3)

Similarly in triangle (A0A4A5) we get
(A0A4) = Sqrt(3)

Hence
(A0A1).(A0A2).(A0A4) = 1.Sqrt(3) Sqrt(3) = 3

Solution 2
Here O is the circumcentre of the triangle (A0A2A4) and
let (A0A2) = a
Then applying basic trigonometry we get:
a = 2RSin(θ)

That is :
a = (2.1.sin(60)) = sqrt(3)

Similarly ,
(A0A4) = Sqrt(3)

Hence
(A0A1).(A0A2).(A0A4) = (1.Sqrt(3) Sqrt(3)) = 3

Solution -3;
This points are the 6th roots of unity. So by applying proeprties of complex no. , we get;
(A0A2) = (Mod{cos(2pi/3)+isin(2pi/3)}-{cos(0) + isin(0)}) = sqrt(3)

Similarly (A0A4) = sqrt(3)
Hence ,
(A0A1).(A0A2).(A0A4) = (1.Sqrt(3) Sqrt(3)) = 3