IGCSE · Practice questions · Right-angled trigonometry: SOH CAH TOA

Label, choose, solve.

Six original Cambridge-style questions on sine, cosine and tangent in right-angled triangles.

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

The method

Label from the angle, then pick the ratio.

Label: hypotenuse first (opposite the right angle), then opposite and adjacent from the angle.
Choose: SOH, CAH or TOA, whichever uses the two sides in the question.
Unknown on top: multiply. Unknown on the bottom: divide.
Angle: work out the ratio, then use \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\).
Accuracy: sides to 3 s.f., angles to 1 d.p.; keep full values between steps.

Calculation

[2 marks]

In triangle \(ABC\), angle \(B = 90^\circ\), \(AC = 12\) cm and angle \(A = 34^\circ\). Calculate \(AB\).

From angle \(A\), \(AB\) is adjacent and \(AC\) is the hypotenuse, so CAH. ✓
\(AB = 12\cos 34^\circ = 9.95\) cm (3 s.f.) ✓

Calculation

[2 marks]

In a right-angled triangle, the side adjacent to an angle of \(48^\circ\) is 7.5 cm. Calculate the length of the hypotenuse.

\(\cos 48^\circ = \dfrac{7.5}{h}\) ✓
\(h = \dfrac{7.5}{\cos 48^\circ} = 11.2\) cm (3 s.f.) ✓

Calculation

[2 marks]

In a right-angled triangle, the two shorter sides are 11 cm and 6.5 cm. Calculate the angle opposite the 11 cm side.

Opposite 11 cm, adjacent 6.5 cm: \(\tan\theta = \dfrac{11}{6.5}\) ✓
\(\theta = \tan^{-1}(1.6923\ldots) = 59.4^\circ\) (1 d.p.) ✓

Analysis

[3 marks]

A building regulation says a wheelchair ramp must make an angle of no more than \(5^\circ\) with the horizontal. A ramp rises 0.5 m over a horizontal distance of 6 m.

Does the ramp meet the regulation? Show your working.

Opposite 0.5 m, adjacent 6 m: \(\tan\theta = \dfrac{0.5}{6}\) ✓
\(\theta = \tan^{-1}(0.08333\ldots) = 4.8^\circ\) (1 d.p.) ✓
\(4.8^\circ \le 5^\circ\), so the ramp meets the regulation. ✓

Calculation

[3 marks]

An isosceles triangle has two sides of 8 cm. The angle between them is \(50^\circ\). Calculate the length of the third side.

The line of symmetry halves the \(50^\circ\) angle and the base, giving a right-angled triangle with hypotenuse 8 cm and angle \(25^\circ\). ✓
Half the base is opposite \(25^\circ\): \(8\sin 25^\circ = 3.3809\ldots\) cm ✓
Base \(= 2 \times 3.3809\ldots = 6.76\) cm (3 s.f.) ✓

Analysis

[4 marks]

A vertical mast \(PQ\) is 6 m tall, with \(Q\) on horizontal ground. A wire joins the top \(P\) to a point \(R\) on the ground and makes an angle of \(58^\circ\) with the ground. A second wire joins \(P\) to a point \(S\) on the ground on the other side of the mast, making an angle of \(40^\circ\) with the ground. \(R\), \(Q\) and \(S\) lie in a straight line.

Calculate the distance \(RS\).

Triangle \(PQR\): \(\tan 58^\circ = \dfrac{6}{QR}\), so \(QR = \dfrac{6}{\tan 58^\circ} = 3.749\ldots\) m ✓
Triangle \(PQS\): \(QS = \dfrac{6}{\tan 40^\circ} = 7.150\ldots\) m ✓
\(RS = QR + QS = 3.749\ldots + 7.150\ldots = 10.899\ldots\) ✓
\(RS = 10.9\) m (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.