Angles

The direct answer is option A: \(\frac{13\pi}{11}\). A reference angle is the small positive angle between the terminal arm and the nearest x-axis. In the third quadrant, the principal angle is found by adding π to the reference angle, because the angle has already passed the negative x-axis. Thus, \(\theta=\pi+\frac{2\pi}{11}=\frac{11\pi}{11}+\frac{2\pi}{11}=\frac{13\pi}{11}\). Option A is correct and lies between π and \(\frac{3\pi}{2}\), the third-quadrant range. Option B, \(\frac{9\pi}{11}\), is less than π, so it lies in the second quadrant. Option C, \(\frac{20\pi}{11}\), is larger than \(\frac{3\pi}{2}\), so it lies in the fourth quadrant rather than the third. Option D, \(\frac{15\pi}{11}\), corresponds to a different reference angle, \(\frac{4\pi}{11}\), not \(\frac{2\pi}{11}\). Memory cue: QIII means π plus the reference angle.

(-725^\circ+1080^\circ=355^\circ). In exams, keep adding multiples of (360^\circ) to a negative angle.

From (A=\frac{1}{2}rs), (r=8), then (\theta=\frac{s}{r}=\frac{25}{8}). In exams, find the radius first from arc length and area.

(15^\circ) per second gives (900^\circ) per minute, and (900^\circ=5\pi) radians. In exams, convert the time unit first.

The angle is (\frac{3\pi}{8}\times10=\frac{15\pi}{4}) radians, which is (675^\circ). In exams, first find the total radians.

(81^\circ=\frac{9\pi}{20}), so (\frac{3x\pi}{20}=\frac{9\pi}{20}) gives (x=3). In exams, write both angles in the same unit.

(-420^\circ30'+720^\circ=299^\circ30'). In exams, add multiples of (360^\circ) even for mixed degree-minute angles.

Direct answer: Option A, \\(\frac{4\pi}{9}\\). Coterminal angles differ by an integer multiple of \\(2\pi\\), so subtract \\(2\pi\\) repeatedly from \\(\frac{58\pi}{9}\\) until the result lies between 0 and \\(2\pi\\). Since \\(2\pi=\frac{18\pi}{9}\\), subtract three such turns: \\(\frac{58\pi}{9}-\frac{54\pi}{9}=\frac{4\pi}{9}\\). This is positive and less than \\(\frac{18\pi}{9}=2\pi\\), so it satisfies the required interval. Option A is correct. Option B, \\(\frac{5\pi}{9}\\), differs from the given angle by \\(\frac{49\pi}{9}\\), not an integer multiple of \\(2\pi\\). Option C, \\(\frac{7\pi}{9}\\), and option D, \\(\frac{8\pi}{9}\\), fail the same coterminal-angle test. The key is not to subtract \\(\pi\\); one full turn is \\(2\pi\\). Memory cue: reduce the coefficient of \\(\pi\\) modulo 2 when angles are written in radians.

The principal angle in the fourth quadrant is (333^\circ), and its greatest negative coterminal angle is (333^\circ-360^\circ=-27^\circ). In exams, keep the greatest negative angle above (-360^\circ).

(36+18\theta=48), so (\theta=\frac{2}{3}) radians. In degrees, this is (\frac{2}{3}\times\frac{180}{\pi}=\frac{120}{\pi}^\circ).

(218^\circ24'=218.4^\circ=\frac{1092}{5}^\circ), so the radian measure is (\frac{91\pi}{75}). In exams, convert (24') into (0.4^\circ).

(54^\circ=\frac{3\pi}{10}) and from (s=r\theta), (r=\frac{9\pi}{\frac{3\pi}{10}}=30). In exams, convert degrees into radians before using the arc formula.

The complementary angle is (\frac{\pi}{2}-\frac{2\pi}{7}=\frac{3\pi}{14}). In degrees, it is (\frac{270}{7}^\circ).

(-\frac{101\pi}{18}+6\pi=\frac{7\pi}{18}). In exams, writing (6\pi) as (\frac{108\pi}{18}) makes addition easy.

(210^\circ=\frac{7\pi}{6}), while (\frac{5\pi}{6}=150^\circ). In exams, convert all options into the same unit before comparing.

The rotation is (-\frac{11}{8}\times2\pi=-\frac{11\pi}{4}), and adding (4\pi) gives (\frac{5\pi}{4}). In exams, first convert revolutions into radians.

The reflex angle is (360^\circ-126^\circ40'=233^\circ20'). In exams, subtract using (1^\circ=60').

The angle is (72^\circ-5\times360^\circ=-1728^\circ). In exams, going behind means subtracting multiples of (360^\circ).

A straight angle is (\pi) radians, so the remaining angle is (\pi-3) radians. In degrees, it is (\frac{180(\pi-3)}{\pi}).

From (s=r\theta), (\theta=\frac{\frac{5\pi r}{6}}{r}=\frac{5\pi}{6}), which is (150^\circ). In exams, cancel (r) to find the angle.

\(\frac{\pi}{3}-(-\frac{5\pi}{3})=2\pi\), 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}).

The direct answer is C, \\(67^\circ30'\\). Let the angle be \\(x\\). Its complement is \\(90^\circ-x\\), and its supplement is \\(180^\circ-x\\). The statement gives \\(90^\circ-x=\frac{1}{5}(180^\circ-x)\\). Multiply by 5: \\(450^\circ-5x=180^\circ-x\\). Rearranging gives \\(270^\circ=4x\\), so \\(x=67.5^\circ=67^\circ30'\\). Option C is correct. Option A, 60°, gives complement 30° and supplement 120°; 30° is not one-fifth of 120° (it would be one-fourth). Option B, 65°, gives 25° and 115°, which do not satisfy the ratio. Option D, 72°, gives 18° and 108°; the ratio is one-sixth, not one-fifth. The angle is between 0° and 90°, as required for a positive complement. Memory cue: write complement as \\(90-x\\) and supplement as \\(180-x\\) before forming the equation.