Angles

(2^\circ30'45''=\frac{1003}{400}^\circ) and the radian value is ( \frac{1003\pi}{72000} ). First convert to degrees and multiply by ( \frac{\pi}{180} ).

( \frac{37\pi}{15}\times\frac{180^\circ}{\pi}=444^\circ). If the denominator is (15), use (180^\circ\div15=12^\circ).

( -\frac{49\pi}{12}+\frac{60\pi}{12}=\frac{11\pi}{12} ). Add enough multiples of (2\pi) to a negative radian angle.

( -2215^\circ+2520^\circ=305^\circ ). Add a suitable multiple of (360^\circ) and then check the range.

( \frac{31\pi}{8}-\frac{16\pi}{8}=\frac{15\pi}{8} ) and the reference angle is (2\pi-\frac{15\pi}{8}=\frac{\pi}{8}). First find the principal angle.

( \frac{19\pi}{7}-2\pi=\frac{5\pi}{7} ), and ( \frac{5\pi}{7} ) lies in the second quadrant. Always reduce the angle before deciding the quadrant.

(1480^\circ-1440^\circ=40^\circ), and it lies in the first quadrant. First find the principal angle for large angles.

( \theta=\frac{s}{r}=\frac{15\pi}{18}=\frac{5\pi}{6}=150^\circ ). First find the radian angle and then convert to degrees.

From ( \frac{81\pi}{8}=\frac{1}{2}\times81\times\theta ), ( \theta=\frac{\pi}{4}=45^\circ ). The angle from the sector area formula is in radians.

Using (A=\frac{1}{2}rs), (121=\frac{1}{2}r\cdot22), so (r=11) and ( \theta=\frac{s}{r}=2 ). When arc and area are given, find (r) first.

The minute hand turns (2\pi) in (60) minutes, so in (18) minutes it turns ( \frac{18}{60}\cdot2\pi=\frac{3\pi}{5} ). Use the time fraction of one full revolution.

At (4:40), the minute hand is at (240^\circ) and the hour hand is at (140^\circ), so the difference is (100^\circ). The hour hand moves (0.5^\circ) per minute.

For coterminal angles the difference can be (360^\circ), so (5x+20^\circ=x+380^\circ) gives (x=90^\circ). Form a linear equation.

(3\theta+45^\circ-(\theta-135^\circ)=360^\circ) gives (2\theta+180^\circ=360^\circ) and ( \theta=90^\circ ). Keep the difference as a multiple of (360^\circ).

Supplementary angles sum to ( \pi ), so ( \pi-\frac{7\pi}{12}=\frac{5\pi}{12} ). In radians use ( \pi ) instead of (180^\circ).

Complementary angles sum to ( \frac{\pi}{2} ), so ( \frac{\pi}{2}-\frac{5\pi}{18}=\frac{2\pi}{9} ). Use a common denominator before subtracting.

( -\frac{23\pi}{10}+\frac{40\pi}{10}=\frac{17\pi}{10} ), which lies between ( \frac{3\pi}{2} ) and (2\pi). First find the principal angle.

( \frac{17\pi}{12} ) lies between ( \pi ) and ( \frac{3\pi}{2} ), and the reference angle is ( \frac{17\pi}{12}-\pi=\frac{5\pi}{12} ). The formula changes by quadrant.

( \frac{5}{3}\times\frac{180^\circ}{3}=100^\circ ). Use the approximation of ( \pi ) given in the question.

The rate is (6) radians per second, so in (3) seconds it turns (18) radians, which is (18\cdot\frac{180^\circ}{\pi}=\frac{3240^\circ}{\pi}).

(225^\circ=\frac{5\pi}{4}) and (r=\frac{s}{\theta}=\frac{25\pi}{5\pi/4}=20). Convert the angle to radians first.

(144^\circ=\frac{4\pi}{5}) and (40\pi=\frac{1}{2}r^2\cdot\frac{4\pi}{5}) gives (r=10). Keep the angle in radians in the area formula.

(2\theta=\frac{22\pi}{9}) and ( \frac{22\pi}{9}-2\pi=\frac{4\pi}{9} ). After multiplying, subtract (2\pi).

( \alpha+\beta=\frac{20\pi}{10}=2\pi ), which is coterminal with (0). In principal angle form, (2\pi) is taken as (0).

(875^\circ-720^\circ=155^\circ), and the reference angle is (180^\circ-155^\circ=25^\circ). Find the principal angle before the reference angle.