Find the area of the region bounded curve y=x^2 the x-axis and lines x=1,x=3
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Sin^2 A = 1/2 tan^2A= 1/(2*(1)^2) =1/2, As angle A is an acute angle.
So,
SinA = 1/√2 , You know value of sin 45°=1/√2
Or,
A = Sin^(-1) {1/√2} = 45°
So, A = 45°
So,
SinA = 1/√2 , You know value of sin 45°=1/√2
Or,
A = Sin^(-1) {1/√2} = 45°
So, A = 45°
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