Angles

One complete revolution measures \(2\pi\) radians. Therefore, the number of revolutions is \(\frac{7\pi}{2}\div 2\pi=\frac{7}{4}\). Hence, the correct answer is \(\frac{7}{4}\) revolutions. \(\frac{3}{2}\) revolutions equals only \(3\pi\) radians, so it is not correct. Exam tip: To convert radians into revolutions, divide the angle by \(2\pi\).

Coterminal angles have the same terminal side, so they differ by an integral multiple of \(360^\circ\). Thus, \(-315^\circ+360^\circ=45^\circ\), and \(45^\circ\) lies in the required interval \(0^\circ\le\theta<360^\circ\). Although \(30^\circ\) is nearby, it does not differ from \(-315^\circ\) by a whole multiple of \(360^\circ\). Exam tip: to express a negative angle in this interval, first add \(360^\circ\).

(4) radians (=4\times\frac{180^\circ}{\pi}\approx229.2^\circ). Use (\pi\approx3.14) for approximation.

For the same arc length (\theta) is inversely proportional to radius. When radius doubles the angle becomes half as (\frac{3\pi}{10}).

One complete revolution measures \(2\pi\) radians. Therefore, \(\frac{3}{8}\) of a revolution is \(\frac{3}{8}\times 2\pi=\frac{3\pi}{4}\) radians. \(\frac{\pi}{2}\) represents only \(\frac{1}{4}\) of a revolution, so it is not correct. Exam tip: To convert a fractional revolution into radians, multiply the fraction by \(2\pi\).

Coterminal angles differ by an integral multiple of \(360^\circ\). Subtract the greatest suitable multiple of \(360^\circ\): \(1440^\circ=4\times360^\circ\). Thus, \(1490^\circ-1440^\circ=50^\circ\). Therefore, the least positive coterminal angle is \(50^\circ\). Although \(40^\circ\) is a close distractor, it does not differ from \(1490^\circ\) by a whole multiple of \(360^\circ\). Exam tip: divide the angle by \(360^\circ\) and take the positive remainder.

In the third quadrant the angle is (\pi+) reference angle. Hence (\pi+\frac{\pi}{7}=\frac{8\pi}{7}).

(405^\circ-45^\circ=360^\circ) so they are coterminal. The difference must be a multiple of (360^\circ).

(1) radian is formed when \(s=r\). For definition based questions remember \(s=r\theta\).

The arc-length formula is \(s=r\theta\), where \(\theta\) must be in radians. \(40^\circ=40\times\frac{\pi}{180}=\frac{2\pi}{9}\) radians. Therefore, \(s=63\times\frac{2\pi}{9}=14\pi\) m. Hence, \(14\pi\) m is correct. A nearby distractor such as \(16\pi\) m can result from an incorrect degree-to-radian conversion or multiplication. Exam tip: always convert the angle to radians before applying the arc-length formula.

Adding (6\pi) to (-\frac{31\pi}{6}) gives (\frac{5\pi}{6}). In exams, add multiples of (2\pi) to negative angles.

(1245^\circ-1080^\circ=165^\circ), so the angle lies in the second quadrant. In exams, first subtract multiples of (360^\circ).

(18^\circ45'=\frac{75}{4}^\circ) and the radian measure is (\frac{75\pi}{720}=\frac{5\pi}{48}). In exams, first convert minutes into degrees.

From (s=r\theta), (\theta=\frac{3\pi}{8}), which is (67.5^\circ). In exams, keep the angle in radians in the arc formula.

(75\pi=\frac{1}{2}\times100\times\theta), so (\theta=\frac{3\pi}{2}=270^\circ). In exams, (\theta) is in radians in the area formula.

(-1540^\circ+1800^\circ=260^\circ), so the principal coterminal angle is (260^\circ). In exams, add multiples of (360^\circ) to negative angles.

(7) revolutions are (14\pi) and (135^\circ=\frac{3\pi}{4}), so the total is (\frac{59\pi}{4}). In exams, convert revolutions and extra angle separately.

The minute hand rotates (2\pi) in (60) minutes, so in (29) minutes it rotates (\frac{29\pi}{30}). In exams, use the time ratio.

(5) hours (36) minutes (=\frac{28}{5}) hours and the hour hand rate is (\frac{\pi}{6}) radians per hour. Hence the angle is (\frac{28}{5}\times\frac{\pi}{6}=\frac{14\pi}{15}).

The principal angle of (\frac{41\pi}{7}) is (\frac{13\pi}{7}), which lies in the fourth quadrant. The reference angle is (2\pi-\frac{13\pi}{7}=\frac{\pi}{7}).