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Chandrashekhar Wankhede 7 years, 10 months ago
Use pythagoras theorem, tangent is perpendicular to radius at point of contact , from given information radius = 3 cm
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Chandrashekhar Wankhede 7 years, 10 months ago
(5π/3) - 2π = -π/3 = -60 degree
Hence cos (5π/3) = cos(-60) = +0.5
Sin (π/6) = +0.5
Hence 3(cos (5π/3)-i Sin (π/6))= 3(0.5 – 0.5i) = 1.5(1-i)
Angle in argand plane = tan-1( y/x) = tan-1( -1/1)= -45 degree = -(π/4)
Modulus of (1-i) = sqrt(2)
Hence 3(cos (5π/3)-i Sin (π/6)) = 1.5 (1-i) = 1.5 * sqrt(2) [cos (-π/4)+isin (-π/4)]
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Chandrashekhar Wankhede 7 years, 10 months ago
Imagine a revolving line in XY plane from origin. The Angle between this line and POSITIVE X AXIS and is measured as POSITIVE when this line is revolved in ANTICLOCKWISE DIRECTION keeping its one end fixed at origin. When revolving line is along POSITIVE X AXIS, angle is zero degree. If line is revolved through 45 degree in anticlockwise direction with respect to positive x axis, line will be in first quadrant, hence angle will be in first quadrant. If line is revolved through 150 degree in anticlockwise direction, angle is in second quadrant. For negative angles line has to be revolved in CLOCKWISE direction. Negative angles are also measured with respect to POSITIVE X AXIS.
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Chandrashekhar Wankhede 7 years, 10 months ago
To verify : n(n+1)(n+5) is divisible by 3 or not
Let us use Principle of Induction
Let f(n) = n(n+1)(n+5) = n3+6n2+5n
Let us verify for n= 1: f(1) = 12 = 3*4 ; Hence f(n) is divisible by 3 for n = 1
Let us assume that f(n) is divisible by 3 for any natural number n = a
That is f(a) = a3+6a2+5a = 3 (m) ;
Let us Prove that f(a+1) is also divisible by 3 if f(a) is divisible by 3
f(a+1) = (a+1)3+6(a+1)2+5(a+1) = a3+9a2+20a+12 = (a3+6a2+5a)+3a2+15a+12 = 3m+3a2+15a+12
= 3(m+a2+5a+4); Hence f(a+1) is also divisible by 3.
Hence it is TRUE that n(n+1)(n+5) is divisible by 3
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