Let A1,A2,....An be the vertices of an n−sided regular polygon such that 1A1A2=1A1A3+1A1A4. Then the value of n is:
A
7
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A7 Let a be the side of n sided regular polygon A1,A2,A3,A4....An ∴ Angle subtended by each side at centre =2πn
Also OA1=OA2=OA3=.....=OAn=r(Say)
In ΔOA1A2,A1A2=2rsinπn
Similarly in A1A3=2rsin2πn
and in A1A4=2rsin3πn
But given that 1A1A2=1A1A3+1A1A4 ⇒12sinπn=12sin2πn+12sin3πn ⇒1sinπn=sin3πn+sin2πnsin2πnsin3πn ⇒1sinπn=sin3πn+sin2πn2sinπncosπnsin3πn ⇒2cosπnsin3πn=sin3πn+sin2πn ⇒sin4πn+sin2πn=sin3πn+sin2πn ∴sin4πn=sin3πn ⇒sin4πn=sin(π−3πn) ⇒4πn=π−3πn ⇒7πn=π
Thus, n=7