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If (-10,8) is a point on the terminal side f a angle 0 in standard position what is the value of tan0?

Sagot :

xero099

Answer:

The value of the tangent is [tex]-\frac{4}{5}[/tex].

Step-by-step explanation:

Let be [tex]A(x,y) = (-10, 8)[/tex] on the terminal side of an angle in standard position and let be [tex]O(x,y)[/tex] the origin of the angle and the origin of the Cartesian plane. By definition of tangent, we make use of the following expression:

[tex]\tan \theta = \frac{y_{A}-y_{O}}{x_{A}-x_{O}}[/tex] (1)

Where:

[tex]x_{A}, x_{O}[/tex] - x-Coordinates of point A and point O.

[tex]y_{A}, y_{O}[/tex] - y-Coordinates of point A and point O.

If we know that [tex]x_{A} = -10, x_{O} = 0[/tex], [tex]y_{A} = 8[/tex] and [tex]y_{O} = 0[/tex], then the value of the tangent is:

[tex]\tan \theta = \frac{y_{A}-y_{O}}{x_{A}-x_{O}}[/tex]

[tex]\tan \theta = \frac{8-0}{-10-0}[/tex]

[tex]\tan \theta = -\frac{4}{5}[/tex]

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