sin θ = 12/13
cos θ = sqrt( 1- sin^2 θ) = sqrt( 1- (12/13)^2 ) = sqrt( 1- 144/169) = sqrt(25)/sqrt(169) = 5/13
Should be shown as
sin ∅ = 12/13
cos ∅ = 5/13
For acute θ, if sinθ = a/b where b ≥ a ≥ 0 (but b > 0), then cosθ = √(b²-a²) / b, because sin²θ + cos²θ = 1
So, cosθ = √(13²-12²) / 13 = 5/13
Did you mean:
sin(θ) = 12/13? If yes, then the adjacent side is: sqrt(13^2-12^2)=5, then:
cos(θ) = 5/13
I assume you mean:
sin θ = opposite / hypotenuse = 12/13
The adjacent leg has to be 5 because:
5² + 12² = 13²
Answer:
cos θ = adjacent / hypotenuse = 5/13
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Answers & Comments
sin θ = 12/13
cos θ = sqrt( 1- sin^2 θ) = sqrt( 1- (12/13)^2 ) = sqrt( 1- 144/169) = sqrt(25)/sqrt(169) = 5/13
Should be shown as
sin ∅ = 12/13
cos ∅ = 5/13
For acute θ, if sinθ = a/b where b ≥ a ≥ 0 (but b > 0), then cosθ = √(b²-a²) / b, because sin²θ + cos²θ = 1
So, cosθ = √(13²-12²) / 13 = 5/13
Did you mean:
sin(θ) = 12/13? If yes, then the adjacent side is: sqrt(13^2-12^2)=5, then:
cos(θ) = 5/13
I assume you mean:
sin θ = opposite / hypotenuse = 12/13
The adjacent leg has to be 5 because:
5² + 12² = 13²
Answer:
cos θ = adjacent / hypotenuse = 5/13