IGCSE · Practice questions · The sine rule

Match each side to its angle.

Extended. Six original Cambridge-style questions on the sine rule in non-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

Find a complete opposite pair first.

Label: side \(a\) opposite angle \(A\), \(b\) opposite \(B\), \(c\) opposite \(C\).
Pair: find a side and its opposite angle (use \(180^\circ\) for the third angle if needed).
Side: \(b = \dfrac{a\sin B}{\sin A}\).
Angle: \(\sin B = \dfrac{b\sin A}{a}\); then ask whether \(180^\circ - B\) also works.
Accuracy: sides 3 s.f., angles 1 d.p.; keep full values between steps.

Calculation

[2 marks]

In triangle \(ABC\), \(A = 38^\circ\), \(B = 71^\circ\) and \(a = 6.4\) cm. Calculate \(b\).

\(\dfrac{b}{\sin 71^\circ} = \dfrac{6.4}{\sin 38^\circ}\) ✓
\(b = \dfrac{6.4\sin 71^\circ}{\sin 38^\circ} = 9.83\) cm (3 s.f.) ✓

Calculation

[3 marks]

In triangle \(ABC\), \(b = 8.3\) cm, \(c = 6.1\) cm and the obtuse angle \(B = 112^\circ\). Calculate angle \(C\).

\(\sin C = \dfrac{6.1\sin 112^\circ}{8.3} = 0.6814\ldots\) ✓
\(C = 43.0^\circ\) (1 d.p.) ✓
\(C\) cannot be \(180^\circ - 43.0^\circ = 137.0^\circ\), since \(B\) is already obtuse. ✓

Calculation

[3 marks]

In triangle \(ABC\), \(A = 55^\circ\), \(B = 48^\circ\) and \(c = 14\) cm. Calculate \(a\).

\(C = 180^\circ - 55^\circ - 48^\circ = 77^\circ\), which pairs with \(c = 14\). ✓
\(a = \dfrac{14\sin 55^\circ}{\sin 77^\circ} = 11.8\) cm (3 s.f.) ✓

Analysis

[4 marks]

In triangle \(ABC\), \(A = 40^\circ\), \(a = 7\) cm and \(b = 10\) cm.

(a) Show that there are two possible values for angle \(B\), and find them.

(b) Find the two possible values of angle \(C\).

(a) \(\sin B = \dfrac{10\sin 40^\circ}{7} = 0.9183\ldots\), so \(B = 66.7^\circ\) or \(B = 180^\circ - 66.7^\circ = 113.3^\circ\). ✓
Both are possible because \(40^\circ + 113.3^\circ = 153.3^\circ < 180^\circ\). ✓
(b) \(C = 180^\circ - 40^\circ - 66.7^\circ = 73.3^\circ\) or \(C = 180^\circ - 40^\circ - 113.3^\circ = 26.7^\circ\). ✓

Proof

[2 marks]

Show that it is not possible to draw a triangle \(ABC\) with \(A = 50^\circ\), \(a = 4\) cm and \(b = 7\) cm.

Using the sine rule, \(\sin B = \dfrac{7\sin 50^\circ}{4} = 1.34\) (3 s.f.). ✓
The sine of an angle cannot exceed 1, so no such angle \(B\) exists, and no such triangle can be drawn. ✓

Analysis

[4 marks]

Two points \(A\) and \(B\) are 50 m apart on one straight bank of a river. A tree \(T\) stands on the opposite bank. Angle \(TAB = 62^\circ\) and angle \(TBA = 75^\circ\).

(a) Calculate the distance \(AT\).

(b) Calculate the width of the river (the perpendicular distance from \(T\) to the line \(AB\)).

(a) Angle \(ATB = 180^\circ - 62^\circ - 75^\circ = 43^\circ\), which faces \(AB = 50\) m. ✓
\(AT\) faces the \(75^\circ\) angle: \(AT = \dfrac{50\sin 75^\circ}{\sin 43^\circ} = 70.8\) m (3 s.f.) ✓
(b) In the right-angled triangle from \(T\) to the bank, \(AT\) is the hypotenuse and the width is opposite the \(62^\circ\): ✓
width \(= 70.815\ldots \times \sin 62^\circ = 62.5\) 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.