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the point `M(-2,9)` lies on the terminal arm of the angle `theta` in standard position....

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yahhhh | Student, Undergraduate | (Level 1) Honors

Posted July 20, 2013 at 3:51 AM via web

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the point `M(-2,9)` lies on the terminal arm of the angle `theta` in standard position. determine the exact value of `tan theta`

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mvcdc | Student, Undergraduate | (Level 1) Associate Educator

Posted July 20, 2013 at 4:24 AM (Answer #1)

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We know that an angle is in standard position if the initial side coincides with the positive x-axis and the vertex is at the origin (0, 0). We can use this to calculate lengths of the sides, and consequenctly the value of the tangent given that the terminal arm contains the point (-2, 9):

 

The point (-2, 9) is 2 units to the left of the y-axis and 9 units upwards of the x-axis. Hence, we can create a right triangle with a short leg of length 2 and a long leg of length 9. [see attached diagram]

 

Then, we know that the tangent can be calculated as:

`tan(theta) = (opposite)/(adjacent)`

Here, the opposite side is the longer leg: 9, and the adjacent side: 2. Also, note that the point is in the second quadrant -- x is negative, y is positive. Since tangent is the ratio of y to x, tangent must also be negative at this quadrant.

Hence, given the above conditions:

`tan(theta) = -9/2`

In general, given a point (x, y) on the terminal side of an angle,

`tan(theta) = y/x`

 

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