Regular Hexagon

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!!

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