On the unit circle, the point P(-5/13, 12/13) lies on the terminal arm of an angle in standard position?

2 Answers
Alan P. · Jim H
Jun 18, 2015

The angle is 1.9656 radians or 112.62^@
(assuming that is what the question intended to ask).

Explanation:

In standard position with the unit length terminal arm at P(-5/13,12/13)
implies
color(white)("XXXX")cos(theta) = -5/13

which, in turn, implies
color(white)("XXXX")theta = arccos(-5/13)
which can be solved using a calculator to obtain the result above.

Jim H
Jun 18, 2015

I have no one sentence answer. See below.

Explanation:

If the question was to find the values of the six trigonometric functions, use the definition. Note that
r = sqrt((-5/13)^2 + (12/13)^2) = sqrt ((25+122)/169) = sqrt (169/169) = 1

(The point P is on the unit circle.)

Call the angle theta (I'll use r in the definitions even though it is 1.

sin theta = y/r = y/1 = 12/13 color(white)"sssss" csc theta = r/y = 1/y = 1/(12/13) = 13/12

cos theta = x/r = x/1 = -5/13 color(white)"sss" sec theta = r/x = 1/x = 1/(-5/13) = -13/5

tan theta = y/x = (12/13)/(-5/13) = 12/13 * -13/5 = -12/5

cot theta = x/y = (-5/13)/(12/13) = -5/13*13/12 = -5/12

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