Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Live 1-1 coding classes to unleash the creator in your Child
let A1, A2, A3 ......An, be the vertices of an n sided regular polygon inscribed in a circle of radius R. If (A1A2)^2+ (A1A3)^2 +.......+ (A1An)^2 = 14(R)^2. Find n.
e know that for a polygon(a = side lenght and n = no of sides), Radius of circumcircle of this polygon =R = a/2sin(pi/n)so a = 2Rsin(pi/n) so A1A2=2Rsinα, A1A3=2Rsin(2α) and A1A4=2Rsin(3α), where α=π/n. now as per given condition1/(2Rsinα) = 1/[2Rsin(2α)] + 1/[2Rsin(3α)] 1/sinα = 1/sin(2α) + 1/sin(3α) 1/sin(2α) = 1/sinα - 1/sin(3α) 1/sin(2α) = [sin(3α) - sinα] / [sinα * sin(3α)] 1/sin(2α) = 2sinα * cos(2α) / [sinα * sin(3α)] 1/sin(2α) = 2cos(2α) /sin(3α) sin(3α) = 2sin(2α)*cos(2α) sin(3α) = sin(4α) 3α + 4α = π 7α = π 7π/n = π n=7
e know that for a polygon(a = side lenght and n = no of sides), Radius of circumcircle of this polygon =
R = a/2sin(pi/n)
so a = 2Rsin(pi/n)
so A1A2=2Rsinα, A1A3=2Rsin(2α) and A1A4=2Rsin(3α), where α=π/n. now as per given condition1/(2Rsinα) = 1/[2Rsin(2α)] + 1/[2Rsin(3α)] 1/sinα = 1/sin(2α) + 1/sin(3α) 1/sin(2α) = 1/sinα - 1/sin(3α) 1/sin(2α) = [sin(3α) - sinα] / [sinα * sin(3α)] 1/sin(2α) = 2sinα * cos(2α) / [sinα * sin(3α)] 1/sin(2α) = 2cos(2α) /sin(3α) sin(3α) = 2sin(2α)*cos(2α) sin(3α) = sin(4α) 3α + 4α = π 7α = π 7π/n = π n=7
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -