Angles

\(\frac{\pi}{3}-(-\frac{5\pi}{3})=2\pi\), so the angles are coterminal and have the same terminal arm. One angle is positive and the other negative, so their rotations are in opposite directions. Exam tip: coterminal angles differ by a multiple of \(2\pi\).

To find a coterminal angle, add or subtract an integer multiple of \(2\pi\). \(-\frac{41\pi}{9}+3(2\pi)=-\frac{41\pi}{9}+\frac{54\pi}{9}=\frac{13\pi}{9}\). Since \(0<\frac{13\pi}{9}<2\pi=\frac{18\pi}{9}\), it lies in the required interval. Although \(\frac{11\pi}{9}\) is close, its difference from the given angle is not an integer multiple of \(2\pi\). Exam tip: use a common denominator to check that the final angle lies between \(0\) and \(2\pi\).

(35^\circ+360^\circ=395^\circ) lies in the given interval. Add or subtract (360^\circ) for coterminal angles.

For fixed arc length (\theta) is inversely proportional to radius. The new radius is (\frac{5}{4}) times so the new angle is (\frac{\pi}{2}).

Perimeter is (2r+s) so \(2r+12=32\) gives \(r=10\). Now \(\theta=\frac{s}{r}=\frac{6}{5}\).

If the angle is (x) then (90^\circ-x=\frac{1}{5}(180^\circ-x)). Solving gives (x=67^\circ30').

The minute hand is at (228^\circ) and the hour hand is at (229^\circ). The smaller difference is (1^\circ).

(125^\circ15'30''=\frac{450930}{3600}^\circ). Multiplying by (\frac{\pi}{180}) gives (\frac{15031\pi}{21600}).

The second hand turns \(2\pi\) in (60) seconds. In (135) seconds the angle is \(\frac{9\pi}{2}\).

Convert \(240^\circ\) into radians: \(240^\circ=\frac{240\pi}{180}=\frac{4\pi}{3}\) radians. Use \(t=\frac{\text{angular displacement}}{\text{angular speed}}\). Thus, \(t=\frac{4\pi/3}{\pi/15}=\frac{4\pi}{3}\times\frac{15}{\pi}=20\) seconds. Therefore, option C is correct. Getting \(24\) seconds indicates an incorrect degree-to-radian conversion. Exam tip: When angular speed is in radians per second, convert the angle to radians first.

One third of the circumference gives one third of the full angle (2\pi). Hence the central angle is (\frac{2\pi}{3}).

The difference (4\theta-120^\circ) must be a multiple of (360^\circ). For the least positive value (\theta=120^\circ).

In the second quadrant the angle is (180^\circ-18^\circ=162^\circ). (162^\circ=\frac{9\pi}{10}) radians.

Let the angle be \(x\) radians. Its complement is \(\frac{\pi}{2}-x\), and its supplement is \(\pi-x\). From the condition, \(\frac{\pi}{2}-x=\frac{1}{3}(\pi-x)\). Multiplying by 3 gives \(\frac{3\pi}{2}-3x=\pi-x\). Hence \(\frac{\pi}{2}=2x\), so \(x=\frac{\pi}{4}\). Thus, option B is correct. For the close distractor \(\frac{\pi}{3}\), the complement is \(\frac{\pi}{6}\) and the supplement is \(\frac{2\pi}{3}\); the complement is \(\frac{1}{4}\), not \(\frac{1}{3}\), of the supplement. Exam tip: use \(\frac{\pi}{2}\) for complement and \(\pi\) for supplement when working in radians.

(18^\circ24'=\frac{92}{5}^\circ). In radians it is (\frac{92}{5}\times\frac{\pi}{180}=\frac{23\pi}{225}).

The angle is negative. Adding \(1440^\circ\), which represents four complete rotations, gives \(-1460^\circ+1440^\circ=-20^\circ\). The angle \(-20^\circ\) is coterminal with \(340^\circ\), and \(340^\circ\) lies between \(270^\circ\) and \(360^\circ\). Hence, its terminal side lies in the fourth quadrant. An angle in the second quadrant would lie between \(90^\circ\) and \(180^\circ\). Exam tip: For a negative angle, add a suitable multiple of \(360^\circ\) to obtain a coterminal angle between \(0^\circ\) and \(360^\circ\).

(\frac{7\pi}{5}) lies in the third quadrant. The reference angle is (\frac{7\pi}{5}-\pi=\frac{2\pi}{5}).

(75^\circ=\frac{5\pi}{12}) and (\frac{53\pi}{12}=4\pi+\frac{5\pi}{12}). (4\pi) contains (2) complete revolutions.

The time difference is (35) minutes. The angle is \(\frac{35}{60}\times2\pi=\frac{7\pi}{6}\) radians.

The time difference is (\frac{11}{2}) hours and the hour hand turns (30^\circ) per hour. The total angle is (165^\circ=\frac{11\pi}{12}).