An angle is the rotation of a ray about its starting point (the vertex). The ray's starting position is the initial side and its final position the terminal side. Rotation anticlockwise gives a positive angle; clockwise gives a negative one. Because the ray can spin past one full turn, there is no upper limit on the size of an angle — $720^\circ$, $-450^\circ$ and $1000^\circ$ are all perfectly valid. Trigonometry needs angles far beyond the $0^\circ$ to $90^\circ$ of a triangle, so we measure rotation itself, in two standard units.
In degree measure one full turn is $360^\circ$, and $1^\circ$ is split into $60$ minutes ($1^\circ = 60'$) and each minute into $60$ seconds ($1' = 60''$). The unit is convenient but arbitrary — there is nothing special about the number $360$; it is simply an old Babylonian choice with many divisors.
In radian measure the angle is the ratio of arc length to radius, which makes it a pure number. One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius:
$$\theta = \dfrac{\text{arc length}}{\text{radius}} = \dfrac{l}{r} \quad\Rightarrow\quad l = r\theta \ \ (\theta \text{ in radians})$$
Since a full circle has circumference $2\pi r$, a complete turn is $\dfrac{2\pi r}{r} = 2\pi$ radians. Equating the two units for a half-turn gives the single conversion fact you must never forget:
$$\pi \text{ radians} = 180^\circ \quad\Rightarrow\quad 1^\circ = \dfrac{\pi}{180}\text{ rad}, \qquad 1 \text{ rad} = \dfrac{180^\circ}{\pi} \approx 57^\circ 16'$$
The common angles, in both units, are worth memorising as a block:
Converting cleanly. To turn degrees into radians, multiply by $\dfrac{\pi}{180}$; to turn radians into degrees, multiply by $\dfrac{180}{\pi}$. When a degree value carries minutes, first convert the minutes to a fraction of a degree using $1' = \left(\dfrac{1}{60}\right)^\circ$, then multiply by $\dfrac{\pi}{180}$. A common slip is to leave $\pi$ in an answer that the question wanted as a decimal, or to forget that a bare number with no degree symbol is already in radians.
Arc length. The defining relation rearranges to the most-used formula of the topic, $l = r\theta$, valid only when $\theta$ is in radians. It says the arc grows in direct proportion both to the radius and to the angle, which is why two circles of different radii subtending equal arcs must have angles in inverse ratio to their radii.
Area of a sector. A sector is the "pizza slice" bounded by two radii and the arc between them. Its area scales with the fraction of the full circle it covers, $\dfrac{\theta}{2\pi}$ of $\pi r^2$, which simplifies neatly:
$$A = \dfrac{\theta}{2\pi}\times \pi r^2 = \dfrac{1}{2} r^2 \theta = \dfrac{1}{2} l r \quad (\theta \text{ in radians})$$
The form $A = \tfrac{1}{2} l r$ is handy when the arc length is already known. Both arc-length and sector-area formulas demand radians — slipping a degree value into them is the single most common source of wrong answers here.
Deeper Insight — why radians, not degrees, are the natural unit: The degree is a human invention; the radian is forced on us by geometry. Because a radian is defined as a ratio of two lengths, it has no units at all, and that single fact is what makes the clean formulas $l = r\theta$ and $A = \tfrac{1}{2}r^2\theta$ possible — neither would hold if $\theta$ were in degrees, since you would have to bury an extra factor of $\tfrac{\pi}{180}$ inside. The deeper payoff comes in calculus: the result $\dfrac{d}{dx}\sin x = \cos x$ is true only when $x$ is in radians, because it secretly relies on $\lim_{\theta \to 0} \tfrac{\sin\theta}{\theta} = 1$, a limit that equals $1$ in radians and $\tfrac{\pi}{180}$ in degrees. So when a question gives no degree symbol, the number is in radians by default. Treat $\pi = 180^\circ$ as a unit conversion, exactly like metres to centimetres, and the whole topic becomes bookkeeping rather than memory.
Convert $40^\circ$ into radian measure.
Solution- Multiply the degree value by $\dfrac{\pi}{180}$.
- $40^\circ = 40 \times \dfrac{\pi}{180} = \dfrac{40\pi}{180}$ rad.
- Simplify the fraction: $\dfrac{40}{180} = \dfrac{2}{9}$.
Answer: $40^\circ = \dfrac{2\pi}{9}$ radians.
Convert $\dfrac{5\pi}{6}$ radians into degree measure.
Solution- Multiply the radian value by $\dfrac{180}{\pi}$.
- $\dfrac{5\pi}{6} \times \dfrac{180}{\pi} = \dfrac{5 \times 180}{6}$.
- $= \dfrac{900}{6} = 150$.
Answer: $\dfrac{5\pi}{6}$ rad $= 150^\circ$.
Find the length of an arc of a circle of radius $14$ cm that subtends an angle of $\dfrac{\pi}{4}$ at the centre.
Solution- Use $l = r\theta$ with $\theta$ already in radians.
- $l = 14 \times \dfrac{\pi}{4} = \dfrac{14\pi}{4} = \dfrac{7\pi}{2}$ cm.
- Taking $\pi \approx \dfrac{22}{7}$: $l \approx \dfrac{7}{2} \times \dfrac{22}{7} = 11$ cm.
Answer: $l = \dfrac{7\pi}{2}$ cm $\approx 11$ cm.
A wheel makes $360$ revolutions in one minute. Through how many radians does it turn in one second?
Solution- $360$ revolutions per minute $= \dfrac{360}{60} = 6$ revolutions per second.
- Each revolution is $2\pi$ radians.
- So in one second it turns $6 \times 2\pi = 12\pi$ radians.
Answer: $12\pi$ radians per second.
The minute hand of a clock is $1.5$ cm long. How far does its tip move in $40$ minutes?
Solution- In $60$ minutes the minute hand sweeps a full turn, $2\pi$ rad, so in $40$ minutes it sweeps $\theta = \dfrac{40}{60}\times 2\pi = \dfrac{4\pi}{3}$ rad.
- The tip traces an arc of radius $r = 1.5$ cm, so $l = r\theta$.
- $l = 1.5 \times \dfrac{4\pi}{3} = 2\pi$ cm.
- Numerically $l \approx 2 \times 3.14 = 6.28$ cm.
Answer: The tip moves $2\pi \approx 6.28$ cm.
Two arcs of the same length subtend angles of $60^\circ$ and $75^\circ$ at the centres of two circles. Find the ratio of their radii.
Solution- Convert: $60^\circ = \dfrac{\pi}{3}$ rad and $75^\circ = \dfrac{5\pi}{12}$ rad.
- Equal arcs means $r_1\theta_1 = r_2\theta_2$, so $\dfrac{r_1}{r_2} = \dfrac{\theta_2}{\theta_1}$.
- $\dfrac{r_1}{r_2} = \dfrac{5\pi/12}{\pi/3} = \dfrac{5\pi}{12}\times\dfrac{3}{\pi} = \dfrac{15}{12} = \dfrac{5}{4}$.
Answer: $r_1 : r_2 = 5 : 4$.
Convert $6$ radians into degree measure (take $\pi = \dfrac{22}{7}$).
Solution- Multiply by $\dfrac{180}{\pi}$: $6 \times \dfrac{180}{\pi} = \dfrac{1080}{\pi}$ degrees.
- Substitute $\pi = \dfrac{22}{7}$: $\dfrac{1080}{22/7} = \dfrac{1080 \times 7}{22} = \dfrac{7560}{22} = \dfrac{3780}{11}$.
- $\dfrac{3780}{11} = 343\dfrac{7}{11}$ degrees; the $\dfrac{7}{11}$ of a degree is $\dfrac{7}{11}\times 60' = 38\dfrac{2}{11}{}'$.
Answer: $6$ rad $= \dfrac{3780}{11}{}^\circ \approx 343^\circ 38'$.
Express $25^\circ 30'$ in radian measure.
Solution- Convert the minutes to degrees: $30' = \left(\dfrac{30}{60}\right)^\circ = (0.5)^\circ$, so the angle is $25.5^\circ = \dfrac{51}{2}{}^\circ$.
- Multiply by $\dfrac{\pi}{180}$: $\dfrac{51}{2}\times\dfrac{\pi}{180} = \dfrac{51\pi}{360}$.
- Simplify by dividing by $3$: $\dfrac{17\pi}{120}$.
Answer: $25^\circ 30' = \dfrac{17\pi}{120}$ radians.
A circle of radius $21$ cm has a sector with central angle $\dfrac{2\pi}{3}$. Find the area of the sector (take $\pi = \dfrac{22}{7}$).
Solution- Use $A = \dfrac{1}{2}r^2\theta$ with $\theta$ in radians.
- $A = \dfrac{1}{2}\times 21^2 \times \dfrac{2\pi}{3} = \dfrac{1}{2}\times 441 \times \dfrac{2\pi}{3} = 147\pi$ cm$^2$.
- Substitute $\pi = \dfrac{22}{7}$: $147 \times \dfrac{22}{7} = 21 \times 22 = 462$ cm$^2$.
Answer: Sector area $= 147\pi \approx 462$ cm$^2$.
The angles of a triangle are in arithmetic progression and the greatest is twice the least. Express the angles in radians.
Solution- Let the angles be $a-d,\ a,\ a+d$ (in degrees). Their sum is $180^\circ$, so $3a = 180^\circ \Rightarrow a = 60^\circ$.
- Greatest $=$ twice least: $a+d = 2(a-d) \Rightarrow 3d = a = 60^\circ \Rightarrow d = 20^\circ$.
- Angles are $40^\circ, 60^\circ, 80^\circ$.
- Convert each by $\times\dfrac{\pi}{180}$: $\dfrac{2\pi}{9},\ \dfrac{\pi}{3},\ \dfrac{4\pi}{9}$.
Answer: The angles are $\dfrac{2\pi}{9},\ \dfrac{\pi}{3},\ \dfrac{4\pi}{9}$ radians (i.e. $40^\circ, 60^\circ, 80^\circ$).
In a circle of radius $5$ cm an arc subtends an angle of $1.2$ radians at the centre. Find the arc length and the area of the corresponding sector.
Solution- Arc length: $l = r\theta = 5 \times 1.2 = 6$ cm.
- Sector area, fastest form: $A = \dfrac{1}{2}lr = \dfrac{1}{2}\times 6 \times 5 = 15$ cm$^2$.
- (Check with $A = \dfrac{1}{2}r^2\theta = \dfrac{1}{2}\times 25 \times 1.2 = 15$ cm$^2$.)
Answer: Arc length $= 6$ cm; sector area $= 15$ cm$^2$.
A railway train runs on a circular track of radius $1500$ m at $66$ km/h. Through what angle (in radians) does it turn in $10$ seconds?
Solution- Convert the speed: $66$ km/h $= \dfrac{66000}{3600} = \dfrac{55}{3}$ m/s.
- Distance (arc) covered in $10$ s: $l = \dfrac{55}{3}\times 10 = \dfrac{550}{3}$ m.
- Angle turned: $\theta = \dfrac{l}{r} = \dfrac{550/3}{1500} = \dfrac{550}{4500} = \dfrac{11}{90}$ rad.
Answer: $\theta = \dfrac{11}{90}$ radians $\approx 0.122$ rad.