AS & A Level · Practice questions · Radians

Think in radians.

Six original Cambridge-style questions on radians: converting between radians and degrees, and solving triangles whose angles are given in radians.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Convert with \(\pi = 180^\circ\), then work in radian mode.

To radians: multiply by \(\dfrac{\pi}{180}\); leave the answer as a multiple of \(\pi\) when it is exact.
To degrees: multiply by \(\dfrac{180}{\pi}\); the \(\pi\) cancels.
Calculator: radian mode for any angle without a degree sign.
Triangles: \(\tfrac{1}{2}ab\sin C\), sine rule and cosine rule as usual; the angles add to \(\pi\).
Accuracy: lengths to 3 s.f., radians to 3 s.f., degrees to 1 d.p.

Recall

[2 marks]

(a) Express \(72^\circ\) in radians, giving your answer in terms of \(\pi\).

(b) Express \(\dfrac{11\pi}{6}\) radians in degrees.

(a) \(72 \times \dfrac{\pi}{180} = \dfrac{72\pi}{180} = \dfrac{2\pi}{5}\) ✓
(b) \(\dfrac{11\pi}{6} \times \dfrac{180}{\pi} = 11 \times 30 = 330^\circ\) ✓

Recall

[2 marks]

(a) Convert \(1.75\) radians to degrees, correct to 1 decimal place.

(b) Convert \(200^\circ\) to radians, correct to 3 significant figures.

(a) \(1.75 \times \dfrac{180}{\pi} = 100.267\ldots = 100.3^\circ\) ✓
(b) \(200 \times \dfrac{\pi}{180} = \dfrac{10\pi}{9} = 3.4906\ldots = 3.49\) rad ✓

Calculation

[4 marks]

In triangle \(ABC\), \(AB = 6\) cm, \(BC = 11\) cm and angle \(ABC = 2.1\) radians.

Find (a) the area of the triangle, (b) the length \(AC\).

(a) Area \(= \tfrac{1}{2} \times 6 \times 11 \times \sin 2.1 = 33 \times 0.86321 = 28.5\ \text{cm}^2\) ✓
(b) \(AC^2 = 36 + 121 - 132\cos 2.1 = 157 - 132 \times (-0.50485) = 223.64\ldots\) ✓
\(AC = 14.95\ldots = 15.0\) cm (3 s.f.) ✓

Calculation

[5 marks]

In triangle \(PQR\), angle \(QPR = \dfrac{\pi}{6}\), angle \(PQR = 0.9\) radians and \(QR = 5\) cm.

(a) Find angle \(PRQ\) in radians.

(b) Find the length \(PQ\).

(a) \(\pi - \dfrac{\pi}{6} - 0.9 = 1.7179\ldots = 1.72\) rad ✓
(b) \(PQ\) is opposite angle \(R\) and \(QR\) is opposite angle \(P\): \(PQ = \dfrac{5\sin 1.7180}{\sin(\pi/6)} = \dfrac{5 \times 0.98919}{0.5} = 9.89\) cm ✓
(c) Angle \(Q\) lies between \(PQ\) and \(QR\): area \(= \tfrac{1}{2} \times 9.8919 \times 5 \times \sin 0.9 = 19.4\ \text{cm}^2\) ✓

Proof

[4 marks]

Triangle \(OAB\) has \(OA = OB = 8\) cm and angle \(AOB = \theta\) radians, where \(\theta\) is obtuse. The area of the triangle is \(16\sqrt{3}\ \text{cm}^2\).

Show that \(\theta = \dfrac{2\pi}{3}\), and find the exact length of \(AB\).

Area \(= \tfrac{1}{2} \times 8 \times 8 \sin\theta = 32\sin\theta = 16\sqrt{3}\), so \(\sin\theta = \dfrac{\sqrt{3}}{2}\). ✓
The solutions between 0 and \(\pi\) are \(\dfrac{\pi}{3}\) and \(\dfrac{2\pi}{3}\); \(\theta\) is obtuse, so \(\theta = \dfrac{2\pi}{3}\). ✓
\(AB = 2 \times 8\sin\dfrac{\pi}{3} = 16 \times \dfrac{\sqrt{3}}{2} = 8\sqrt{3}\) cm ✓

Analysis

[5 marks]

A clock has a minute hand of length 12 cm and an hour hand of length 8 cm, both pivoted at the centre of the face.

(a) Find, in radians in terms of \(\pi\), the angle between the hands at exactly 2 o'clock.

(b) Find the exact distance between the tips of the hands at 2 o'clock.

(a) Each hour mark is \(\dfrac{2\pi}{12} = \dfrac{\pi}{6}\), so at 2 o'clock the angle is \(2 \times \dfrac{\pi}{6} = \dfrac{\pi}{3}\). ✓
(b) \(d^2 = 12^2 + 8^2 - 2 \times 12 \times 8\cos\dfrac{\pi}{3} = 208 - 96 = 112\), so \(d = \sqrt{112} = 4\sqrt{7}\) cm. ✓
(c) At 4 o'clock the angle is \(\dfrac{2\pi}{3}\): \(d^2 = 208 - 192\cos\dfrac{2\pi}{3} = 208 + 96 = 304\). ✓
\(d = \sqrt{304} = 17.4\) cm (3 s.f.) ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.