Coterminal angles: keep the landing, change the journey

You are changing the recorded turn by complete laps while keeping the same final ray. Find a positive angle larger than 360° coterminal with 30°.

1. 30° + 360° = 390°.A full revolution returns to the same direction before the last 30° turn.

You are changing the recorded turn by complete laps while keeping the same final ray. Find the angle in 0° ≤ θ < 360° coterminal with −75°.

1. −75° + 360° = 285°.Adding a full revolution turns the clockwise measurement into a nonnegative one with the same finish.
2. Stop at 285°.0° ≤ 285° < 360°, so it is inside the requested lap.

You are changing the recorded turn by complete laps while keeping the same final ray. Reduce 405° to a coterminal measurement in 0° ≤ θ < 360°.

1. 405° − 360° = 45°.Remove one whole lap while preserving the terminal side.
2. 0° ≤ 45° < 360°, so stop at 45°.The remainder is in the requested one-lap range; subtracting another 360° would make it negative.

You are changing the recorded turn by complete laps while keeping the same final ray. Reduce 1230° to a coterminal measurement in 0° ≤ θ < 360°.

1. 3 × 360° = 1080°.Three full turns can be removed without changing the terminal side.
2. 1230° − 1080° = 150°.This leaves the remainder after the three complete laps.
3. 0° ≤ 150° < 360°, so stop at 150°.The remainder is inside one lap; that is the range the question requests.

You are finding a positive journey and a negative journey to the same ray, each shorter than one revolution. The positive turn is one and one quarter right angles. Give one positive and one negative measure between −360° and 360°.

1. One and one quarter is 54$\frac{5}{4}$, so the positive measure is 54$\frac{5}{4}$ × 90° = 5 × 22.5° = 112.5°.A right angle measures 90°. Each quarter of it is 90° ÷ 4 = 22.5°, and five such quarters total 112.5°.
2. 112.5° − 360° = −247.5°.Subtract one full turn to get a negative journey with the same initial and terminal sides.

A technician closes a water valve by turning its handwheel clockwise through 1777.5°. Viewed from where the technician stands, place the center of the wheel at the origin of a coordinate grid, with the positive x-axis pointing right and the positive y-axis pointing up. A painted spoke starts along the positive x-axis, and that ray is the initial side of every angle in this problem. Counterclockwise turns are positive and clockwise turns are negative. (a) Write the spoke's rotation as a signed angle. Then find the angle θ with 0° ≤ θ < 360° that is coterminal with it, and describe where the spoke stops. (b) Find two positive angles and two negative angles, other than the rotation itself, that are coterminal with the spoke's rotation.

1. Fix the initial side on the positive x-axis, where the painted spoke starts, and write the clockwise turn as the signed angle −1777.5°.Coterminal angles must share the initial side as well as the terminal side. An angle measured from a different starting ray could stop on the same ray without being coterminal. Counterclockwise is the positive direction, so a clockwise turn gets a negative sign.
2. Add 360° one full turn at a time: −1777.5° → −1417.5° → −1057.5° → −697.5° → −337.5° → 22.5°.The measure is negative, so keep adding 360° until it is at least 0°. A full turn changes the journey but not the ray where the spoke stops. After four turns the measure is still negative (−337.5°). The fifth turn gives 22.5°, the first measure that is at least 0°.
3. Write the five turns as one calculation: −1777.5° + 360° × 5 = −1777.5° + 1800° = 22.5°, so θ = 22.5°.This is −1777.5° + 360°n with the integer n = 5. As a shortcut, 360° × 4 = 1440° is less than 1777.5° but 360° × 5 = 1800° is more, so five is the fewest full turns that bring the measure to at least 0°. Since 22.5° is already less than 360°, no turn needs to be subtracted.
4. Locate the landing: 22.5° = 14$\frac{1}{4}$ × 90°, so the spoke stops 14$\frac{1}{4}$ of a right angle counterclockwise from the positive x-axis, in Quadrant I.1777.5° is 22.5° less than five full clockwise turns (1800°). So the spoke halts 22.5° before it would sweep back down onto the positive x-axis. That is the ray at 22.5°, which agrees with the calculation.
5. Positive answers: keep θ = 22.5° and add one more turn: 22.5° + 360° = 382.5°.Both measures are positive and distinct. Each differs from −1777.5° by whole turns, so each stops on the same ray.
6. Negative answers: subtract full turns from 22.5°: 22.5° − 360° = −337.5°, then −337.5° − 360° = −697.5°.Each subtraction is one more clockwise lap, which keeps the stopping ray. Both results are negative, distinct, and not the original −1777.5°. They also appear as stops in the chain of additions above.