I won't provide a drawing, but here's the points! (Bold is the ratios)
To draw a vector to the point (-3, 9), we need to plot a line segment from the origin (0, 0) to the point (-3, 9).
To draw a vector to the point (-3, 9), we need to plot a line segment from the origin (0, 0) to the point (-3, 9).
Here's a visual representation:
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--------|------(-3, 9)
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(0, 0)
Now, let's calculate the six trigonometric ratios for the right triangle formed by the vector:
Sine (sinθ) = Opposite/Hypotenuse
In this case, the opposite side is 9 units (vertical distance) and the hypotenuse is the length of the vector, which can be calculated using the Pythagorean theorem:
Hypotenuse = sqrt((-3)^2 + 9^2) = sqrt(9 + 81) = sqrt(90)
Therefore, sinθ = 9/sqrt(90).
Cosine (cosθ) = Adjacent/Hypotenuse
In this case, the adjacent side is -3 units (horizontal distance), and the hypotenuse is sqrt(90) as calculated above.
Therefore, cosθ = -3/sqrt(90).
Tangent (tanθ) = Opposite/Adjacent
In this case, the opposite side is 9 units (vertical distance), and the adjacent side is -3 units (horizontal distance).
Therefore, tanθ = 9/-3 = -3.
Cosecant (cscθ) = 1/sinθ
Therefore, cscθ = 1/(9/sqrt(90)) = sqrt(90)/9.
Secant (secθ) = 1/cosθ
Therefore, secθ = 1/(-3/sqrt(90)) = -sqrt(90)/3.
Cotangent (cotθ) = 1/tanθ
Therefore, cotθ = 1/(-3) = -1/3.
These are the six trigonometric ratios for the right triangle formed by the vector to the point (-3, 9).