If cos(θ) = 4/9 with θ in Quadrant I, what is sin(θ)?

Answer 1

cos(θ) = 4/9 ← given

sin(θ)² + cos(θ)² = 1

sin(θ) = √1 - (4/9)² ≅ 0.802

(Angle ≅ 53.3°)

Answer 2

9^2 - 4^2 = 81 - 16 = 65

sin(theta) = sqr65/9 which is slightly more than 8/9