Angles

\(225^\circ=\frac{5\pi}{4}\) radians. \(r=\frac{s}{\theta}=\frac{15\pi}{5\pi/4}=12\) cm.

The perimeter of a sector is \(2r+s\), where \(s\) is the arc length. From the condition, \(2r+s=3s\). Thus, \(2r=2s\), so \(s=r=8\) cm. Using \(s=r\theta\), we get \(\theta=\frac{s}{r}=\frac{8}{8}=1\) radian. At \(2\) radians, the arc length would be \(16\) cm, so the perimeter would not be three times the arc length. Exam tip: Use \(s=r\theta\) only when \(\theta\) is measured in radians.

Coterminal angles differ by an integral multiple of 360°. \(-1025^\circ+3\times360^\circ=-1025^\circ+1080^\circ=55^\circ\). Therefore, the least positive coterminal angle is 55°. Choosing 45° or 65° would not give a difference that is a whole multiple of 360°. Exam tip: For a negative angle, keep adding 360° until the result lies between 0° and 360°.

Subtract a multiple of \(2\pi\) to obtain a coterminal angle: \(\frac{31\pi}{6}-4\pi=\frac{31\pi}{6}-\frac{24\pi}{6}=\frac{7\pi}{6}\). Since \(\pi<\frac{7\pi}{6}<\frac{3\pi}{2}\), the terminal side lies in the third quadrant. It is \(\frac{\pi}{6}\) past \(\pi\), so the second quadrant is not correct. Exam tip: First reduce a radian angle to a coterminal angle between \(0\) and \(2\pi\).

Coterminal angles differ by an integer multiple of \(2\pi\). Subtracting \(4\pi=\frac{32\pi}{8}\) from \(\frac{5\pi}{8}\) gives \(\frac{5\pi}{8}-\frac{32\pi}{8}=-\frac{27\pi}{8}\). Moreover, \(-4\pi=-\frac{32\pi}{8}< -\frac{27\pi}{8}< -\frac{16\pi}{8}=-2\pi\), so option D lies in the required interval. Option A differs from \(\frac{5\pi}{8}\) by \(-3\pi\), which is not an integer multiple of \(2\pi\). Exam tip: write coterminal angles as \(\theta+2n\pi\) and then test the required interval.

The point is in the second quadrant and the reference angle is (\frac{\pi}{6}). Hence the principal angle is (\frac{5\pi}{6}).

The fourth-quadrant ray of (y=-x) has reference angle (\frac{\pi}{4}). Hence the principal angle is (\frac{7\pi}{4}).

(11) revolutions are (3960^\circ) and (\frac{2\pi}{3}=120^\circ). The total angle is (4080^\circ).

On the negative (x)-axis the angles are \(\pi\) and \(3\pi\). Hence \(m=18,54\) and the sum is (72).

A right angle is (\frac{\pi}{2}) so the reference angle is (\frac{\pi}{8}). In the second quadrant the angle is (\frac{7\pi}{8}).

The area ratio is (\frac{\theta}{2\pi}). From (\frac{\theta}{2\pi}=\frac{3}{4}) we get (\theta=\frac{3\pi}{2}).

(-75^\circ15'=-\frac{301}{4}^\circ). In radians it is (-\frac{301}{4}\times\frac{\pi}{180}=-\frac{301\pi}{720}).

The difference is (5x+60^\circ) and it must be a multiple of (360^\circ). For the least positive value (x=60^\circ).

Coterminal angles differ by \(2k\pi\), where \(k\) is an integer. Hence, \(\frac{5\pi}{12}+\frac{n\pi}{6}=\frac{17\pi}{12}+2k\pi\). Subtracting \(\frac{5\pi}{12}\) gives \(\frac{n\pi}{6}=\pi+2k\pi\), so \(n=6+12k\). The least positive value is obtained for \(k=0\), giving \(n=6\). With \(n=5\), the difference is \(\frac{5\pi}{6}\), not an integral multiple of \(2\pi\). Exam tip: test coterminality by checking whether the angle difference is an integral multiple of \(2\pi\).

After (m) minutes the angle difference is (90^\circ-5.5m). For the next right angle (m=\frac{360}{11}).

The arc-length formula is \(s=r\theta\), where \(\theta\) must be measured in radians. Hence, \(\frac{s}{r}=\theta\). Since \(150^\circ=150\times\frac{\pi}{180}=\frac{5\pi}{6}\), the required ratio is \(\frac{5\pi}{6}\). Note that \(\frac{2\pi}{3}\) corresponds to \(120^\circ\), not \(150^\circ\). Exam tip: In arc-length questions, convert the angle to radians first.

(5) radians (=5\times\frac{180^\circ}{\pi}). Using (\pi=\frac{22}{7}) gives approximately (286^\circ21'49'').

The difference of coterminal angles is an integer multiple of (360^\circ). (\frac{1980}{360}=\frac{11}{2}) is not an integer.

The perimeter is (24+12\theta) and the area is (72\theta). From (24+12\theta=72\theta) we get (\theta=\frac{2}{5}).

The minute hand is at (324^\circ) and the hour hand is at (357^\circ). The smaller difference is (33^\circ).