Its center at O = (0, 0), the unit circle x2 + y2 = 1 goes through P = (1, 0). The line y = 0.6 intersects the circle at A and B, with A in the first quadrant. The angles POA and POB are said to be in standard position, because their initial ray OP points in the positive x-direction. (Their terminal rays are OA and OB.) Find the sizes of these angles.

Question

How are they related?

Then (Continuation) If we restrict ourselves to a single revolution, there are actually two angles in standard position that could be named POB. The one determined by minor arc PB is said to be positive, because it opens in the counterclockwise direction. Find its degree measure. The one determined by major arc PB is said to be negative, because it opens in the clockwise direction. Find its degree measure.

Answer 1

To find the coordinates of point A, we plug y=0.6 into the equation of the unit circle:

x^2 + (0.6)^2 = 1

x^2 + 0.36 = 1

x^2 = 0.64

x = ±0.8

So point A has coordinates (0.8, 0.6) (since it's in the first quadrant) and point B has coordinates (-0.8, 0.6).

Using the coordinates of point A, we can find the cosine and sine of angle POA:

cos(POA) = 0.8

sin(POA) = 0.6

To find the angle POA, we can use the inverse cosine function:

POA = arccos(0.8) ≈ 36.87°

Similarly, we can find the angle POB:

cos(POB) = -0.8

sin(POB) = 0.6

POB = arccos(-0.8) ≈ 143.13°

The angles POA and POB are related because they are supplementary, meaning their sum is equal to 180°.

For the positive angle POB, the degree measure is 143.13° (as found above) since it refers to the angle determined by minor arc PB and opens in the counterclockwise direction.

For the negative angle POB, we need to find the angle determined by major arc PB and opening in the clockwise direction. To do this, we can simply subtract the positive angle POB from 360°:

Negative angle POB = 360° - 143.13° ≈ 216.87° (in negative direction)