Chapter 3 · Trigonometric Functions

3.1 Angles: Degree and Radian Measure

An angle is an amount of turning — it can be negative, and it can exceed a full turn. We measure it in degrees or in radians, where one radian is the angle that cuts off an arc exactly one radius long, so that $s=r\theta$.

In Class 10 every angle lived inside a triangle, so it was always between $0^\circ$ and $180^\circ$. But a ceiling fan blade turns through thousands of degrees every minute, the minute hand of a clock turns clockwise while a screw being loosened turns the other way, and a spin bowler's wrist may rotate through more than a full turn before the ball is released. To describe turning, an angle has to be allowed to be as large as we like, and to carry a sign that tells us the direction.

In this section we define angles as rotations, measure them in degrees and in a second unit, the radian, and learn to convert between the two. Radians may look strange at first, but they are the natural unit of mathematics: with them, the length of an arc is simply radius × angle, and — as you will see in Chapter 12 — the formulae of calculus for sine and cosine only take their simple form when angles are in radians.

Angles as Rotations

📘 Definition 3.1 — Angle; positive and negative angles

An angle is the amount of rotation of a ray about its end point. The starting position of the ray is the initial side, its final position is the terminal side, and the point about which it turns is the vertex.

If the rotation is anticlockwise the angle is positive; if it is clockwise the angle is negative. An angle is in standard position when its vertex is at the origin and its initial side lies along the positive $x$-axis; it then "lies in" the quadrant that contains its terminal side. An angle whose terminal side lies along a coordinate axis is a quadrantal angle. Angles with the same initial and terminal sides — such as $30^\circ$, $390^\circ$ and $-330^\circ$ — are called coterminal; they differ by a whole number of complete turns.

📘 Definition 3.2 — Degree measure

If the rotation from the initial side to the terminal side is $\left(\tfrac{1}{360}\right)$th of a complete revolution, the angle has a measure of one degree, written $1^\circ$. A degree is divided into $60$ minutes and a minute into $60$ seconds:

$$1^\circ=60',\qquad 1'=60''.$$

So one complete anticlockwise revolution is $360^\circ$, a right angle is $90^\circ$, and $1''=\left(\tfrac{1}{3600}\right)^\circ$.

Because a rotation can go on for more than one turn, $720^\circ$ (two anticlockwise turns) and $-450^\circ$ (one and a quarter clockwise turns) are perfectly good angles. To find the quadrant of a large angle, add or subtract multiples of $360^\circ$ until you reach an angle between $0^\circ$ and $360^\circ$ — this does not move the terminal side.

✏️ Example 1 — Rotations, signs and quadrants

(a) Find the angle, in degrees, for $\tfrac34$ of a revolution anticlockwise, and for $2\tfrac12$ revolutions clockwise.
(b) Through what angle does the hour hand of a clock turn between 3 o'clock and 7 o'clock?
(c) In which quadrant does the terminal side of each angle lie: $400^\circ$, $-135^\circ$, $1000^\circ$?

💡 Show solution▶

(a) $\tfrac34\times360^\circ=270^\circ$, positive because the turn is anticlockwise. $2\tfrac12\times360^\circ=900^\circ$, clockwise, so the angle is $-900^\circ$.

(b) The hour hand covers $360^\circ$ in $12$ hours, i.e. $30^\circ$ per hour. From 3 to 7 is $4$ hours: $4\times30^\circ=120^\circ$, turned clockwise. The angle is $-120^\circ$.

(c) $400^\circ=360^\circ+40^\circ$: coterminal with $40^\circ$, first quadrant. $-135^\circ+360^\circ=225^\circ$: third quadrant. $1000^\circ=2\times360^\circ+280^\circ$: coterminal with $280^\circ$, fourth quadrant.

Check $-135^\circ$ means turning clockwise past the negative $y$-axis ($-90^\circ$) but not as far as the negative $x$-axis ($-180^\circ$) — between them lies the third quadrant ✓.

✏️ Example 2 — Degrees, minutes and seconds

(a) Express $40.35^\circ$ in degrees and minutes. (b) Express $25^\circ15'36''$ in degrees as a decimal.

💡 Show solution▶

(a) The whole part is $40^\circ$. The decimal part: $0.35^\circ=0.35\times60'=21'$. So $40.35^\circ=40^\circ21'$.

(b) $15'=\left(\tfrac{15}{60}\right)^\circ=0.25^\circ$ and $36''=\left(\tfrac{36}{3600}\right)^\circ=0.01^\circ$. So $25^\circ15'36''=25.26^\circ$.

Check $0.26^\circ\times60=15.6'$, and $0.6'\times60=36''$ — we get back $15'36''$ ✓.

Radian Measure

Degrees divide a full turn into $360$ parts — a choice made by Babylonian astronomers, not by the geometry of the circle. A more natural way to measure an angle at the centre of a circle is to ask: how long is the arc it cuts off, compared with the radius?

📘 Definition 3.3 — Radian

The angle subtended at the centre of a circle by an arc whose length equals the radius of the circle is one radian, written $1$ rad (or $1^{\text{c}}$). More generally, if an arc of length $s$ in a circle of radius $r$ subtends the angle $\theta$ at the centre, the radian measure of the angle is

$$\theta=\frac{s}{r}\ \text{radians}.$$

A clockwise angle gets a negative radian measure. When no unit is written — as in $\sin 2$ or "an angle of $\tfrac{\pi}{3}$" — the angle is in radians.

Why is this a fair definition? In two circles of different sizes, sectors with the same central angle are similar figures: if the radius is doubled, the arc is doubled too. So the ratio $s/r$ depends only on the angle, not on the size of the circle — the radian is the same angle in every circle.

📐 Theorem 3.1 — Arc length

In a circle of radius $r$, an arc that subtends an angle of $\theta$ radians at the centre has length

$$s=r\theta .$$

Proof. In a fixed circle, the length of an arc is proportional to the angle it subtends at the centre (doubling the angle doubles the arc). An angle of $1$ radian cuts off an arc of length $r$, by Definition 3.3; so an angle of $\theta$ radians cuts off an arc of length $\theta\times r$. $\blacksquare$

The formula needs $\theta$ in radians. If the angle is given in degrees, convert it first.

📐 Theorem 3.2 — Relation between radians and degrees

$$\pi\ \text{radians}=180^\circ .$$

Hence $1\ \text{rad}=\left(\dfrac{180}{\pi}\right)^\circ\approx57.30^\circ$ (about $57^\circ17'45''$; with $\pi\approx\tfrac{22}{7}$ it comes out as $57^\circ16'$), and $1^\circ=\dfrac{\pi}{180}\ \text{rad}\approx0.01746\ \text{rad}$.

Proof. A complete revolution cuts off the whole circumference, an arc of length $2\pi r$. By Theorem 3.1 the angle of one revolution is $\dfrac{2\pi r}{r}=2\pi$ radians. But one revolution is also $360^\circ$. So $2\pi\ \text{rad}=360^\circ$, i.e. $\pi\ \text{rad}=180^\circ$. $\blacksquare$

🧭 Problem-solving strategy — Converting between degrees and radians

Degrees → radians: multiply by $\dfrac{\pi}{180}$. First turn minutes and seconds into a decimal (or a fraction) of a degree.
Radians → degrees: multiply by $\dfrac{180}{\pi}$. If $\pi$ does not cancel, use the value of $\pi$ you are told ($\tfrac{22}{7}$ or $3.14$) and turn the decimal part of the answer into minutes and seconds.
Keep answers like $\tfrac{5\pi}{12}$ as exact multiples of $\pi$ unless a decimal is asked for.

degrees	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$180^\circ$	$270^\circ$	$360^\circ$
radians	$0$	$\tfrac{\pi}{6}$	$\tfrac{\pi}{4}$	$\tfrac{\pi}{3}$	$\tfrac{\pi}{2}$	$\tfrac{2\pi}{3}$	$\pi$	$\tfrac{3\pi}{2}$	$2\pi$

✏️ Example 3 — Degrees to radians

Convert into radian measure: (a) $240^\circ$ (b) $40^\circ20'$ (c) $-47^\circ30'$.

💡 Show solution▶

(a) $240\times\dfrac{\pi}{180}=\dfrac{4\pi}{3}$ rad.

(b) $20'=\left(\tfrac{20}{60}\right)^\circ=\left(\tfrac13\right)^\circ$, so $40^\circ20'=\left(\tfrac{121}{3}\right)^\circ$. Then $\dfrac{121}{3}\times\dfrac{\pi}{180}=\dfrac{121\pi}{540}$ rad $\approx0.704$ rad.

(c) $47^\circ30'=47.5^\circ=\left(\tfrac{95}{2}\right)^\circ$, so $-47^\circ30'=-\dfrac{95}{2}\times\dfrac{\pi}{180}=-\dfrac{19\pi}{72}$ rad.

Check (a): $\tfrac{4\pi}{3}=\pi+\tfrac{\pi}{3}$, i.e. $180^\circ+60^\circ=240^\circ$ ✓.

✏️ Example 4 — Radians to degrees

Convert into degree measure: (a) $\dfrac{5\pi}{3}$ (b) $\dfrac{11}{16}$ rad (c) $-4$ rad. In (b) and (c) take $\pi=\tfrac{22}{7}$ and give the answer in degrees, minutes and seconds.

💡 Show solution▶

(a) $\dfrac{5\pi}{3}\times\dfrac{180}{\pi}=300^\circ$ — here $\pi$ cancels, so no approximation is needed.

(b) $\dfrac{11}{16}\times\dfrac{180}{\pi}=\dfrac{11}{16}\times\dfrac{180\times7}{22}=\dfrac{315}{8}=39\tfrac38$ degrees. Now $\tfrac38^\circ=\tfrac38\times60'=22\tfrac12'$ and $\tfrac12'=30''$. Answer: $39^\circ22'30''$.

(c) $-4\times\dfrac{180\times7}{22}=-\dfrac{2520}{11}=-229\tfrac{1}{11}$ degrees. $\tfrac1{11}^\circ=\tfrac{60}{11}'=5\tfrac{5}{11}'$, and $\tfrac{5}{11}'=\tfrac{300}{11}''\approx27''$. Answer: about $-229^\circ5'27''$.

Check (c): $1$ rad $\approx57.3^\circ$, so $-4$ rad $\approx-229.2^\circ$ ✓ (the small difference comes from $\tfrac{22}{7}$ being slightly larger than $\pi$).

Arc Length and Its Applications

The arc is cut into pieces, each exactly one radius long (alternating colours). The number of radius-lengths in the arc is the angle in radians. On the right the arc is straightened out next to the radius. Change the radius: the arc grows, but the number of radius-lengths — the angle — stays the same. A full turn needs $2\pi\approx6.28$ of them.

radius $r$:angle:

$\theta$ (radians)—

$\theta$ (degrees)—

radius $r$—

arc $s=r\theta$—

✏️ Example 5 — The tip of a minute hand

The minute hand of a wall clock is $1.5$ cm long. How far does its tip move in $40$ minutes? (Use $\pi=3.14$.)

💡 Show solution▶

Angle The minute hand turns once ($2\pi$ rad) in $60$ minutes. In $40$ minutes it turns $\tfrac{40}{60}\times2\pi=\tfrac{4\pi}{3}$ rad.

Arc The tip moves along a circle of radius $r=1.5$ cm, so $s=r\theta=1.5\times\tfrac{4\pi}{3}=2\pi$ cm.

Answer $s=2\pi\approx6.28$ cm.

Check In a full hour the tip travels the whole circumference $2\pi(1.5)=3\pi\approx9.42$ cm; $40$ minutes is $\tfrac23$ of that, $6.28$ cm ✓.

✏️ Example 6 — Same arc, different circles; a swinging pendulum

(a) In two circles, arcs of the same length subtend angles of $60^\circ$ and $75^\circ$ at the centres. Find the ratio of their radii.
(b) A pendulum $75$ cm long swings so that its bob moves along an arc of length $10$ cm. Find the angle through which it swings, in radians and (approximately) in degrees.

💡 Show solution▶

(a) Convert: $60^\circ=\tfrac{\pi}{3}$, $75^\circ=\tfrac{5\pi}{12}$. Equal arcs: $r_1\cdot\tfrac{\pi}{3}=r_2\cdot\tfrac{5\pi}{12}$, so $\dfrac{r_1}{r_2}=\dfrac{5\pi/12}{\pi/3}=\dfrac54$. The radii are in the ratio $5:4$ — the circle with the smaller angle is the bigger circle.

(b) The bob moves on a circle of radius $75$ cm: $\theta=\dfrac{s}{r}=\dfrac{10}{75}=\dfrac{2}{15}$ rad.

Degrees $\dfrac{2}{15}\times\dfrac{180}{\pi}=\dfrac{24}{\pi}\approx7.64^\circ$ — about $7^\circ38'$.

Check (a): with $r_1=5$, $r_2=4$, the arcs are $5\cdot\tfrac{\pi}{3}=\tfrac{5\pi}{3}$ and $4\cdot\tfrac{5\pi}{12}=\tfrac{5\pi}{3}$ — equal ✓.

⚠️ Common mistakes with angle measure

1. Using degrees in $s=r\theta$. A $60^\circ$ arc of radius $5$ cm is not $300$ cm long! Convert first: $5\times\tfrac{\pi}{3}\approx5.24$ cm.

2. Calculator in the wrong mode. $\sin 30$ in RAD mode is $\sin(30\text{ rad})\approx-0.988$, not $0.5$. Check the DEG/RAD indicator before every trigonometry calculation.

3. "$\pi=180$". $\pi$ is the number $3.14159\ldots$; it is $\pi$ radians that equals $180^\circ$. Writing $\pi=180$ leads to nonsense such as $\sin\pi=\sin180$.

4. Minutes are not decimals. $40^\circ30'$ is $40.5^\circ$, not $40.3^\circ$: divide the minutes by $60$.

5. Forgetting the sign. A clock hand turns clockwise, so as a rotation its angle is negative. When only a distance or a size is asked for, give the magnitude.

Exercises

Angles as rotations; degrees, minutes and seconds.

1 Find the angle in degrees for $\tfrac16$ of a revolution anticlockwise.
2 Find the angle in degrees for $3$ revolutions clockwise.
3 Find the angle in degrees for $\tfrac{5}{12}$ of a revolution clockwise.
4 In which quadrant does the terminal side of $520^\circ$ lie?
5 In which quadrant does the terminal side of $-300^\circ$ lie?
6 Find the angle between $0^\circ$ and $360^\circ$ that is coterminal with $-750^\circ$, and name its quadrant.
7 Express $63.45^\circ$ in degrees and minutes.
8 Express $12^\circ24'36''$ in degrees as a decimal.
9 Through what angle (with sign) does the hour hand turn from 12 noon to 4:30 pm?

💡 Show solutions▶

1 $\tfrac16\times360^\circ=60^\circ$.

2 $3\times360^\circ=1080^\circ$ clockwise: $-1080^\circ$.

3 $\tfrac5{12}\times360^\circ=150^\circ$ clockwise: $-150^\circ$.

4 $520^\circ-360^\circ=160^\circ$: second quadrant.

5 $-300^\circ+360^\circ=60^\circ$: first quadrant.

6 $-750^\circ+3\times360^\circ=330^\circ$: fourth quadrant.

7 $0.45\times60'=27'$: $63^\circ27'$.

8 $24'=0.4^\circ$, $36''=0.01^\circ$: $12.41^\circ$.

9 $4.5$ hours $\times30^\circ$ per hour $=135^\circ$, clockwise: $-135^\circ$.

Converting between degrees and radians.

1 Convert $25^\circ$ into radians.
2 Convert $520^\circ$ into radians.
3 Convert $75^\circ$ into radians.
4 Convert $39^\circ22'30''$ into radians.
5 Convert $-225^\circ$ into radians.
6 Convert $\dfrac{7\pi}{6}$ into degrees.
7 Convert $-\dfrac{3\pi}{4}$ into degrees.
8 Convert $6$ rad into degrees, minutes and seconds (take $\pi=\tfrac{22}{7}$).
9 Convert $-2$ rad into degrees, minutes and seconds (take $\pi=\tfrac{22}{7}$).
10 The angles of a triangle are in the ratio $2:3:4$. Find them in radians.

Check: convert any angle between degrees, radians and D°M′S″

degrees:39.375°

D°M′S″—

radians—

as a multiple of $\pi$—

💡 Show solutions▶

1 $25\times\tfrac{\pi}{180}=\tfrac{5\pi}{36}$ rad.

2 $520\times\tfrac{\pi}{180}=\tfrac{26\pi}{9}$ rad.

3 $\tfrac{5\pi}{12}$ rad.

4 $39^\circ22'30''=39.375^\circ=\left(\tfrac{315}{8}\right)^\circ$; $\tfrac{315}{8}\times\tfrac{\pi}{180}=\tfrac{7\pi}{32}$ rad (compare Example 4(b)).

5 $-\tfrac{5\pi}{4}$ rad.

6 $\tfrac76\times180^\circ=210^\circ$.

7 $-\tfrac34\times180^\circ=-135^\circ$.

8 $6\times\tfrac{180\times7}{22}=\tfrac{3780}{11}=343\tfrac{7}{11}$ degrees; $\tfrac7{11}^\circ=38\tfrac{2}{11}'$ and $\tfrac2{11}'\approx11''$: about $343^\circ38'11''$.

9 $-2\times\tfrac{1260}{22}=-\tfrac{1260}{11}=-114\tfrac{6}{11}$ degrees; $\tfrac6{11}^\circ=32\tfrac{8}{11}'$ and $\tfrac8{11}'\approx44''$: about $-114^\circ32'44''$.

10 The angles sum to $\pi$: $2k+3k+4k=\pi$ gives $k=\tfrac{\pi}{9}$, so the angles are $\tfrac{2\pi}{9}$, $\tfrac{\pi}{3}$, $\tfrac{4\pi}{9}$ (that is $40^\circ$, $60^\circ$, $80^\circ$).

III

Arc length $s=r\theta$. Use $\pi=\tfrac{22}{7}$ where a decimal is needed.

1 Find the arc length cut off by an angle of $2.5$ rad in a circle of radius $20$ cm.
2 Find the length of an arc of a circle of radius $7$ cm that subtends $60^\circ$ at the centre.
3 An arc of length $15$ cm lies on a circle of radius $6$ cm. Find the angle it subtends, in radians and approximately in degrees.
4 An arc of length $22$ cm subtends $30^\circ$ at the centre. Find the radius of the circle.
5 A chord of a circle of radius $21$ cm is equal in length to the radius. Find the angle it subtends at the centre and the length of the minor arc.
6 The minute hand of a watch is $2.1$ cm long. How far does its tip move in $20$ minutes?
7 A pendulum $50$ cm long swings through an angle of $18^\circ$. Find the length of the arc described by its bob.
8 A wheel makes $300$ revolutions per minute. Through how many radians does it turn in one second?
9 In two circles, arcs of the same length subtend $45^\circ$ and $60^\circ$ at the centres. Find the ratio of the radii.

💡 Show solutions▶

1 $s=20\times2.5=50$ cm.

2 $s=7\times\tfrac{\pi}{3}=\tfrac{7\pi}{3}=\tfrac{22}{3}\approx7.33$ cm.

3 $\theta=\tfrac{15}{6}=2.5$ rad $\approx2.5\times57.30^\circ\approx143.2^\circ$.

4 $30^\circ=\tfrac{\pi}{6}$, so $r=\dfrac{s}{\theta}=\dfrac{22}{\pi/6}=\dfrac{132}{\pi}=132\times\tfrac{7}{22}=42$ cm.

5 The chord and two radii form an equilateral triangle, so the angle is $60^\circ=\tfrac{\pi}{3}$. Minor arc: $21\times\tfrac{\pi}{3}=7\pi=22$ cm.

6 $\theta=\tfrac{20}{60}\times2\pi=\tfrac{2\pi}{3}$; $s=2.1\times\tfrac{2\pi}{3}=1.4\pi=4.4$ cm.

7 $18^\circ=\tfrac{\pi}{10}$; $s=50\times\tfrac{\pi}{10}=5\pi\approx15.7$ cm.

8 $300$ rev/min $=5$ rev/s; each revolution is $2\pi$ rad, so $10\pi\approx31.4$ rad in one second.

9 $r_1\cdot\tfrac{\pi}{4}=r_2\cdot\tfrac{\pi}{3}$, so $r_1:r_2=\tfrac13:\tfrac14=4:3$.

Applications and harder problems.

1 Find, in degrees and in radians, the smaller angle between the hands of a clock at 3:40.
2 Cycling. A bicycle wheel of diameter $70$ cm turns through $5$ radians without slipping. How far does the bicycle move?
3 Geography. The equatorial radius of the Earth is about $6378$ km. Two towns on the equator have longitudes differing by $1^\circ$. How far apart are they along the equator?
4 Railways. A train runs at $66$ km/h along a circular curve of radius $1500$ m. Through what angle (in radians and roughly in degrees) does it turn in $10$ s?
5 The perimeter of a sector of a circle of radius $r$ is equal to half the circumference of the circle. Find the angle of the sector in radians and approximately in degrees.
6 Explain, using similar sectors, why the angle "one radian" does not depend on the size of the circle used to define it.

💡 Show solutions▶

1 Measured clockwise from 12: the minute hand is at $40\times6^\circ=240^\circ$; the hour hand at $3\times30^\circ+40\times0.5^\circ=110^\circ$. The difference is $130^\circ=\tfrac{13\pi}{18}$ rad (less than $180^\circ$, so it is the smaller angle).

2 Radius $35$ cm: $s=35\times5=175$ cm $=1.75$ m. Rolling without slipping means the ground covered equals the arc turned.

3 $s=6378\times\tfrac{\pi}{180}\approx111.3$ km.

4 $66$ km/h $=\tfrac{66000}{3600}\approx18.33$ m/s, so in $10$ s the train covers about $183.3$ m. $\theta=\tfrac{183.3}{1500}\approx0.122$ rad $\approx7.0^\circ$.

5 Perimeter $=2r+r\theta$; half the circumference $=\pi r$. So $2+\theta=\pi$, $\theta=\pi-2\approx1.14$ rad $\approx65.4^\circ$.

6 Two sectors with the same central angle in circles of radii $r$ and $R$ are similar, so their arcs are in the ratio $r:R$. If the arc of the first equals $r$, the arc of the second equals $R$. So "arc = radius" picks out the same angle in both circles.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE Main

How this is examined: Board papers ask for conversions (1–2 marks) and short arc-length problems on clocks, wheels and pendulums (2–3 marks), often as a case study. In JEE, radian measure is never asked on its own — but every later trigonometry question assumes you can move between degrees and radians instantly and use $s=r\theta$ without a slip.

E1. Board · 1 mark The radian measure of $-37^\circ30'$ is (A) $-\dfrac{5\pi}{24}$ (B) $-\dfrac{7\pi}{24}$ (C) $-\dfrac{5\pi}{12}$ (D) $-\dfrac{\pi}{5}$

💡 Show solution▶

Idea $37^\circ30'=37.5^\circ=\left(\tfrac{75}{2}\right)^\circ$, and $\tfrac{75}{2}\times\tfrac{\pi}{180}=\tfrac{75\pi}{360}=\tfrac{5\pi}{24}$.

Trap (C) is $75^\circ$ — forgetting to halve. Reading $30'$ as $0.3^\circ$ gives $37.3^\circ$, which matches none of the options.

Answer (A) $-\dfrac{5\pi}{24}$.

E2. Board · 2 marks A pendulum $60$ cm long swings so that its bob describes an arc of $12$ cm. Find the angle of swing in radians and in degrees.

💡 Show solution▶

Radians $\theta=\dfrac{s}{r}=\dfrac{12}{60}=\dfrac15$ rad. (1 mark)

Answer In degrees: $\dfrac15\times\dfrac{180}{\pi}=\dfrac{36}{\pi}\approx11.46^\circ$, about $11^\circ28'$. (1 mark)

E3. Board · 3 marks Find, in degrees and radians, the angle between the hour hand and the minute hand of a clock at 7:20.

💡 Show solution▶

Minute hand It turns $6^\circ$ per minute: at 20 minutes it is $120^\circ$ clockwise from 12. (1 mark)

Hour hand It turns $30^\circ$ per hour and $0.5^\circ$ per minute: at 7:20 it is $7\times30^\circ+20\times0.5^\circ=220^\circ$ from 12. (1 mark)

Answer The angle between them is $220^\circ-120^\circ=100^\circ=\dfrac{5\pi}{9}$ rad. (1 mark)

E4. Board · Case study A merry-go-round at a fair has radius $4.2$ m and makes one complete revolution every $24$ s. Take $\pi=\tfrac{22}{7}$. (i) Through what angle, in radians, does it turn in $10$ s? (2 marks) (ii) How far does a child on the outer edge travel in those $10$ s? (1 mark) (iii) Another child sits $2.1$ m from the centre. How far does she travel in the same time? (1 mark)

💡 Show solution▶

(i) One revolution, $2\pi$ rad, takes $24$ s, so in $10$ s the angle is $\tfrac{10}{24}\times2\pi=\tfrac{5\pi}{6}$ rad. (2 marks)

(ii) $s=4.2\times\tfrac{5\pi}{6}=3.5\pi=3.5\times\tfrac{22}{7}=11$ m. (1 mark)

Answer (iii) Same angle, half the radius: $s=2.1\times\tfrac{5\pi}{6}=5.5$ m — half as far, though both children turn through the same angle. (1 mark)

E5. JEE Main The angles of a quadrilateral are in arithmetic progression with common difference $10^\circ$. The largest angle, in radians, is (A) $\dfrac{7\pi}{12}$ (B) $\dfrac{5\pi}{12}$ (C) $\dfrac{2\pi}{3}$ (D) $\dfrac{3\pi}{4}$

💡 Show solution▶

Idea Angles $a$, $a+10^\circ$, $a+20^\circ$, $a+30^\circ$ add up to $360^\circ$: $4a+60^\circ=360^\circ$, so $a=75^\circ$.

Trap (B) is $75^\circ$, the smallest angle. (D) $=135^\circ$ comes from using a common difference of $30^\circ$.

Answer (A): the largest is $105^\circ=105\times\tfrac{\pi}{180}=\tfrac{7\pi}{12}$.

E6. JEE Main Numerical value. A cycle wheel of radius $35$ cm rolls along a straight road without slipping. Taking $\pi=\tfrac{22}{7}$, find the number of complete revolutions the wheel makes while the cycle travels $1.1$ km.

💡 Show solution▶

Idea In one revolution the wheel turns $2\pi$ rad and moves $s=r\cdot2\pi=2\times\tfrac{22}{7}\times35=220$ cm.

Count $1.1$ km $=110000$ cm, and $\tfrac{110000}{220}=500$.

Answer $500$. ⏱ Fast route: angle turned $=\tfrac{s}{r}=\tfrac{110000}{35}$ rad; divide by $2\pi=\tfrac{44}{7}$ to get $500$ turns.

🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. Is one radian really the same angle in every circle — and how close can you get to $57.3^\circ$ with a piece of string?

Why it matters. Engineers specify the speed of motors, turbines and hard disks in radians per second, and satellite tracking stations convert arcs of an orbit into angles with $s=r\theta$. A unit that is independent of the size of the circle is what makes these conversions work.

What to do.

Collect four round objects of different sizes (a bangle, a steel plate, a bucket, a bicycle wheel). Measure each diameter and halve it to get $r$.
Cut a string of length $r$, lay it along the rim, and mark its ends. Join both marks to the centre and measure the angle with a protractor.
Repeat three times for each object and record the mean angle and its spread.
Lay the string round the rim as many times as it will go. How many lengths fit in a full turn? Compare with $2\pi\approx6.28$.

Tools: String, ruler or measuring tape, protractor, spreadsheet

Hand in: A table of $r$ and measured angles, the overall mean compared with $\tfrac{180}{\pi}^\circ$, and a sentence on the biggest source of error.

Going further. A ceiling fan's blades are $60$ cm long and the fan runs at $300$ revolutions per minute. Find the angular speed in rad/s and the speed of a blade tip in m/s.

Instructor note: expect angles of $55^\circ$–$60^\circ$; the main error is locating the centre of the object. For the fan: $300$ rpm $=10\pi\approx31.4$ rad/s, and the tip speed is $0.6\times10\pi\approx18.8$ m/s.

🧪 Self-check quiz

0/5 correct

Q1. One radian is approximately

Q2. The radian measure of $150^\circ$ is

Q3. An arc of length $12$ cm on a circle of radius $8$ cm subtends at the centre an angle of

Q4. The terminal side of $-200^\circ$ lies in the

Q5. The size of the angle (in radians) through which the minute hand of a clock turns in $25$ minutes is

← Chapter 3 — Trigonometric Functions 3.2 Trigonometric Functions of a Real Number →