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Draw a vector to the point (-3,9). Find the six trigonometric ratios for that right triangle.

User Phil Kulak
by
7.9k points

1 Answer

6 votes

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).

User Noel Frostpaw
by
9.0k points

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