In a right triangle ABD, right angled at B, < BAC= theta , AD= b, BD= a, c is the mid-point of BD. Show that Sin2 theta+ cos2 theta = 1
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Sinθ = BC/AC & Sin²θ = BC²/AC² and Cosθ = AB/AC & Cos²θ = AB²/AC²
right angled ∆ ABC
∠ABC = ∠B = 90°
∠BAC = ∠A = θ
Draw a line segment AB
Draw an angle of 90 deg at B on AB using setsquare / protector
Draw an acute angle (let say θ) at A on AB using protector such that it intersect angle drawn earlier at C
ΔABC is constructed
AB = Base
BC = Perpendicular
AC = hypotenuse
Sinθ = Perpendicular / Hypotenuse
=> Sinθ = BC/AC
Sin²θ = BC²/AC²
Cosθ = Base / Hypotenuse
=> Cosθ = AB/AC
Cos²θ = AB²/AC²
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