Negative sqrt(2) is.
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Tan(Pi/5) = √(5-2*√(5)) ~= 0.7265
The area of a circle with radius 5 is 25 pi. Concentric circles with radius 3 and 4 have areas of 9 pi and 16 pi. The concentric circle with radius four consumes the circle with radius 3. 25 pi minus 16 pi leaves 9 pi of the circle with radius 5 left over. 16 pi is slightly over three-fifths of the circle with radius 5.
Pi / 5 would be in Quadrant I.
1.25
7*pi/5 = 1.4*pi pi < 1.4*pi < 1.5*pi and so the angle is in the third quadrant.