Angles

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

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 \(360^\circ\).

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

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}).

One complete revolution is \\(360^\\circ\\). Therefore, 11 complete revolutions produce \\(11\\times360^\\circ=3960^\\circ\\). The additional angle is \\(2\\pi/3\\) radians. Since \\(\\pi\\) radians equals \\(180^\\circ\\), this angle equals \\(2\\times180^\\circ/3=120^\\circ\\). Adding both parts gives \\(3960^\\circ+120^\\circ=4080^\\circ\\). Hence option C is correct.

The important point is that revolutions and radians must first be expressed in the same unit before they are added. The 11 turns are not 11 degrees; each full turn contributes 360 degrees. Similarly, \\(2\\pi/3\\) is one-third of a full half-turn, or 120 degrees. The total angle is therefore 4080 degrees, not 3960 degrees, which would omit the extra rotation. The supplied answer C and explanation are accurate.

A point lies on the negative x-axis when its angle is an odd multiple of \\(\\pi\\): \\(\\pi,3\\pi,5\\pi,\\ldots\\). The condition here is \\(0<\\theta<4\\pi\\), so only \\(\\theta=\\pi\\) and \\(\\theta=3\\pi\\) are included. Substituting \\(\\theta=m\\pi/18\\), we get \\(m\\pi/18=\\pi\\), giving \\(m=18\\), and \\(m\\pi/18=3\\pi\\), giving \\(m=54\\). Their sum is \\(18+54=72\\), so option C is correct.

The strict inequalities matter: zero is excluded, and \\(4\\pi\\) itself is also excluded. Although \\(5\\pi\\) is again on the negative x-axis, it is outside the stated interval. Angles differing by \\(2\\pi\\) point in the same direction, which explains the two permitted positions, \\(\\pi\\) and \\(3\\pi\\), within this interval. Thus the supplied answer C is correct and the explanation identifies all possible values without adding an invalid endpoint.

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 \(2\pi\).

The direct answer is C: \\(\frac{360}{11}\\) minutes after 3 o’clock. At exactly 3:00, the hands are 90° apart. After \\(m\\) minutes, the minute hand moves \\(6m^\circ\\), while the hour hand moves \\(0.5m^\circ\\). Thus the separation decreases at \\(6-0.5=5.5^\circ\\) per minute, so the separation is \\(90-5.5m\\). For the first later right angle, set it equal to 90° after the hands have passed through the other orientation, or equivalently use the relative movement needed to gain 90°: \\(5.5m=180^\circ\\), giving \\(m=180/5.5=360/11\\) minutes. Option C is correct, about 32 minutes 44 seconds after 3, near 3:32:44. Option A, \\(300/11\\), and B, \\(330/11\\), do not produce a 90° separation. Option D, \\(390/11\\), is later than the first occurrence. Common mistake: using only the minute-hand speed and forgetting that the hour hand also moves.

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.