NCERT Class 11 Mathematics — Trigonometric Functions, Exercise 3.1. All 7 questions solved.

The chapter starts by replacing degrees with radians, and this exercise is entirely about that change of units and the one formula it makes possible.

$$\pi \text{ radian} = 180^\circ \quad\Longrightarrow\quad 1^\circ = \frac{\pi}{180} \text{ radian}, \qquad 1 \text{ radian} = \frac{180}{\pi} \text{ degrees}$$

$$\boxed{\ l = r\theta\ } \qquad (\theta \text{ in radians})$$

For the sub-degree units, $1^\circ = 60’$ (minutes) and $1′ = 60”$ (seconds).

Key insight. $l = r\theta$ works only when $\theta$ is in radians, and that is the entire reason radians exist. It also means the formula can be run in any direction: given two of $l$, $r$, $\theta$, the third follows. Questions 4, 5 and 7 find $\theta$; question 6 compares two circles by eliminating $l$ between two copies of the same equation.

Question 1

Find the radian measures corresponding to the following degree measures:

(i) $25^\circ$    (ii) $-47^\circ 30’$    (iii) $240^\circ$    (iv) $520^\circ$

Solution. Multiply each by $\dfrac{\pi}{180}$ and reduce.

(i)

$$25^\circ = 25 \times \frac{\pi}{180} = \frac{25\pi}{180} = \frac{5\pi}{36} \text{ radian}$$

(ii) Convert the minutes to a decimal (or a fraction) first: $30′ = \tfrac{30}{60}^\circ = \tfrac12 ^\circ$, so the angle is $-47\tfrac12^\circ = -\tfrac{95}{2}^\circ$.

$$-\frac{95}{2} \times \frac{\pi}{180} = -\frac{95\pi}{360} = -\frac{19\pi}{72} \text{ radian}$$

(iii)

$$240^\circ = \frac{240\pi}{180} = \frac{4\pi}{3} \text{ radian}$$

(iv)

$$520^\circ = \frac{520\pi}{180} = \frac{26\pi}{9} \text{ radian}$$

Note that $520^\circ$ is more than a full turn; there is no need to reduce it modulo $360^\circ$ unless the question asks for a coterminal angle.

Question 2

Find the degree measures corresponding to the following radian measures (use $\pi = \dfrac{22}{7}$):

(i) $\dfrac{11}{16}$    (ii) $-4$    (iii) $\dfrac{5\pi}{3}$    (iv) $\dfrac{7\pi}{6}$

Solution. Multiply each by $\dfrac{180}{\pi}$. Where $\pi$ does not cancel, use $\dfrac{180}{\pi} = 180 \times \dfrac{7}{22} = \dfrac{630}{11}$.

(i)

$$\frac{11}{16} \times \frac{630}{11} = \frac{630}{16} = 39.375^\circ$$

Now convert the decimal part into minutes and seconds. $0.375^\circ = 0.375 \times 60′ = 22.5’$, and $0.5′ = 30”$:

$$39.375^\circ = 39^\circ\,22’\,30”$$

(ii)

$$-4 \times \frac{630}{11} = -\frac{2520}{11} = -229.0909\ldots^\circ$$

The fractional part is $\tfrac{1}{11}^\circ$. Then $\tfrac{1}{11} \times 60′ = 5.4545\ldots’$, so $5’$ with $0.4545\ldots’ \times 60 \approx 27”$:

$$-4 \text{ radian} \approx -229^\circ\,5’\,27”$$

(iii) Here $\pi$ cancels outright, so no approximation is needed:

$$\frac{5\pi}{3} \times \frac{180}{\pi} = 5 \times 60 = 300^\circ$$

(iv)

$$\frac{7\pi}{6} \times \frac{180}{\pi} = 7 \times 30 = 210^\circ$$

Question 3

A wheel makes $360$ revolutions in one minute. Through how many radians does it turn in one second?

Solution. Convert the rate to revolutions per second first, then revolutions to radians.

$$360 \text{ revolutions per minute} = \frac{360}{60} = 6 \text{ revolutions per second}$$

One complete revolution is $2\pi$ radians, so

$$6 \times 2\pi = 12\pi \text{ radian per second}$$

Question 4

Find the degree measure of the angle subtended at the centre of a circle of radius $100$ cm by an arc of length $22$ cm (use $\pi = \dfrac{22}{7}$).

Solution. From $l = r\theta$,

$$\theta = \frac{l}{r} = \frac{22}{100} \text{ radian}$$

Converting to degrees:

$$\theta = \frac{22}{100} \times \frac{180}{\pi} = \frac{22}{100} \times 180 \times \frac{7}{22} = \frac{1260}{100} = 12.6^\circ$$

The $22$ cancels neatly against $\pi = \tfrac{22}{7}$, which is why the question specifies that value. Finally, $0.6^\circ = 0.6 \times 60′ = 36’$:

$$\theta = 12^\circ\,36’$$

Question 5

In a circle of diameter $40$ cm, the length of a chord is $20$ cm. Find the length of the minor arc of the chord.

Solution. The diameter is $40$ cm, so the radius is $r = 20$ cm. The chord is also $20$ cm — equal to the radius.

Join the two ends of the chord to the centre. The two radii and the chord form a triangle with all three sides equal to $20$ cm, so it is equilateral and the angle at the centre is

$$\theta = 60^\circ = \frac{\pi}{3} \text{ radian}$$

The minor arc is the one cut off by this angle, so

$$l = r\theta = 20 \times \frac{\pi}{3} = \frac{20\pi}{3} \text{ cm}$$

Recognising the equilateral triangle is the whole question — no trigonometric ratio is needed at all.

Question 6

If in two circles, arcs of the same length subtend angles $60^\circ$ and $75^\circ$ at the centre, find the ratio of their radii.

Solution. Let the radii be $r_1$ and $r_2$ and convert both angles to radians:

$$\theta_1 = 60^\circ = \frac{\pi}{3}, \qquad \theta_2 = 75^\circ = \frac{5\pi}{12}$$

The arcs have the same length $l$, so

$$l = r_1\theta_1 = r_2\theta_2 \quad\Longrightarrow\quad \frac{r_1}{r_2} = \frac{\theta_2}{\theta_1}$$

$$\frac{r_1}{r_2} = \frac{5\pi/12}{\pi/3} = \frac{5\pi}{12} \times \frac{3}{\pi} = \frac{5}{4}$$

$$r_1 : r_2 = 5 : 4$$

Note the ratio comes out inverted relative to the angles — the circle that turns through the smaller angle must have the larger radius to sweep out the same arc.

Question 7

Find the angle in radian through which a pendulum swings if its length is $75$ cm and the tip describes an arc of length

(i) $10$ cm    (ii) $15$ cm    (iii) $21$ cm

Solution. The pendulum’s length is the radius of the arc its tip traces, so $r = 75$ cm and $\theta = \dfrac{l}{r}$ in each case.

(i)

$$\theta = \frac{10}{75} = \frac{2}{15} \text{ radian}$$

(ii)

$$\theta = \frac{15}{75} = \frac{1}{5} \text{ radian}$$

(iii)

$$\theta = \frac{21}{75} = \frac{7}{25} \text{ radian}$$

All three answers are pure numbers, with no $\pi$ in them — that is normal for radians, which are a ratio of two lengths and therefore dimensionless.

Common mistakes

• Question 1(ii), treating $30’$ as $0.30^\circ$. Minutes are sixtieths, so $30′ = 0.5^\circ$. The same slip in reverse spoils question 2 and question 4.
• Question 2, converting the decimal remainder wrongly. After $\tfrac{630}{16} = 39.375^\circ$, the fractional part must be multiplied by $60$ to become minutes, and the fractional part of that by $60$ again to become seconds.
• Question 2(iii) and (iv), substituting $\pi = \tfrac{22}{7}$ unnecessarily. When the angle already contains $\pi$, it cancels exactly and the answer is a whole number of degrees. Substituting the approximation introduces a rounding error into an exact result.
• Question 3, forgetting the “per second”. $360$ revolutions per minute has to be divided by $60$ before converting to radians. Answering $720\pi$ is the result of skipping that step.
• Question 5, using the diameter as the radius. $r = 20$ cm, not $40$. And the chord equalling the radius is the clue that makes the triangle equilateral — miss it and there is no route to the angle.
• Question 6, writing $r_1 : r_2 = 4 : 5$. The radii are inversely proportional to the angles, so the larger angle belongs to the smaller circle.
• Using $l = r\theta$ with $\theta$ in degrees. The formula is a radian identity. In question 4, using $\theta = 12.6$ directly would give an arc length of $1260$ cm rather than $22$ cm.

Practise next

• Exercise 3.2 — the trigonometric functions themselves, evaluated for angles in every quadrant and for angles beyond one full turn.
• Exercise 3.3 — the sum and difference identities, where the radian measures of question 1 become the standard arguments $\tfrac{\pi}{6}$, $\tfrac{\pi}{4}$, $\tfrac{\pi}{3}$.