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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
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