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Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative.

1. \(\alpha=60^{\circ}\)
2. \(\beta=-225^{\circ}\)
3. \(\gamma=540^{\circ}\)
4. \(\varphi=-750\)
in Mathematics by Platinum (132,156 points) | 166 views

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1. To graph \(\alpha=60^{\circ}\), we draw an angle with its initial side on the positive \(x\)-axis and rotate counter-clockwise \(\frac{60^{\circ}}{360^{\circ}}=\frac{1}{6}\) of a revolution. We see that \(\alpha\) is a Quadrant I angle.
To find angles which are coterminal, we look for angles \(\theta\) of the form \(\theta=\alpha+360^{\circ} \cdot k\) for some integer \(k\).
- When \(k=1\), we get \(\theta=60^{\circ}+360^{\circ}=420^{\circ}\).
- Substituting \(k=-1\) gives \(\theta=60^{\circ}-360^{\circ}=-300^{\circ}\).
- If we let \(k=2\), we get \(\theta=60^{\circ}+720^{\circ}=780^{\circ}\).


2. Since \(\beta=-225\) is negative, we start at the positive \(x\)-axis and rotate clockwise \(\frac{225}{360^{\circ}}=\frac{5}{8}\) of a revolution. We see that \(\beta\) is a Quadrant II angle.

To find coterminal angles, we proceed as before and compute \(\theta=-225^{\circ}+360^{\circ} \cdot k\) for integer values of \(k\). Letting \(k=1\)
\(k=-1\) and \(k=2\), we find \(135^{\circ},-585^{\circ}\) and \(495^{\circ}\) are all coterminal with \(-225^{\circ}\).

3. Since \(\gamma=540\) is positive, we rotate counter-clockwise from the positive \(x\)-axis. One full revolution accounts for \(360^{\circ}\), with \(180^{\circ}\), or half of a revolution, remaining. Since the terminal side of \(\gamma\) lies on the negative \(x\)-axis, \(\gamma\) is a quadrantal angle.

All angles coterminal with \(\gamma\) are of the form \(\theta=540^{\circ}+360^{\circ} \cdot k\), where \(k\) is an integer. Working through the arithmetic, we find three such angles: \(900^{\circ}, 180^{\circ}\) and \(-180^{\circ}\).

4. The Greek letter \(\varphi\) is pronounced 'fee' or 'fie'9 and since \(\varphi=-750 \quad\) is negative, we begin our rotation clockwise from the positive \(x\)-axis. Two full rotations account for \(720^{\circ}\), with just \(30^{\circ}\) or \(\frac{1}{12}\) of a revolution to go.
We find that \(\varphi\) is a Quadrant IV angle. To find coterminal angles, we compute \(\theta=-750^{\circ}+360^{\circ} \cdot k\) for a few integers \(k\) and obtain \(-390^{\circ},-30^{\circ}\) and \(330^{\circ} .\)

by Platinum (132,156 points)

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