Extended. SOH CAH TOA only works with a right angle. The sine rule works in every triangle: each side divided by the sine of its opposite angle gives the same number.
In any triangle \(ABC\), with side \(a\) opposite angle \(A\) and so on: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\). Use it when you know a side AND its opposite angle. To find an angle, flip it: \(\dfrac{\sin A}{a} = \dfrac{\sin B}{b}\).
Fig. 1: triangle ABC drawn to scale with A = 40°, B = 65°, C = 75° and a = 8 cm. Each side is named after the angle opposite it: b = 11.3 cm and c = 12.0 cm
Section 01
Opposite pairs.
In the lab, change the angles of the triangle and watch the three ratios \(\dfrac{a}{\sin A}\), \(\dfrac{b}{\sin B}\), \(\dfrac{c}{\sin C}\). However you stretch the triangle, all three stay equal to each other.
Naming: capital letters for angles, the matching small letter for the side opposite: side \(a\) faces angle \(A\).
When to use it: you know one complete opposite pair (a side and the angle facing it), plus one more side or angle.
Finding a side: put the sides on top, \(\dfrac{b}{\sin B} = \dfrac{a}{\sin A}\), so \(b = \dfrac{a\sin B}{\sin A}\).
Finding an angle: put the sines on top, \(\dfrac{\sin B}{b} = \dfrac{\sin A}{a}\), so \(\sin B = \dfrac{b\sin A}{a}\).
Missing pair? If you know two angles, the third is \(180^\circ\) minus the other two. That often completes a pair.
Worked example Finding a side
In Fig. 1, \(A = 40^\circ\), \(B = 65^\circ\) and \(a = 8\) cm. Find \(b\).
The known pair is \(a = 8\) with \(A = 40^\circ\). We want \(b\), opposite \(B = 65^\circ\).
Answer: \(c = 12.0\) cm (3 s.f.). The longest side faces the largest angle, as it should.
Worked example Finding an angle
In triangle \(ABC\), \(a = 9\) cm, \(b = 7\) cm and \(A = 70^\circ\). Find angle \(B\).
Sines on top: \(\dfrac{\sin B}{7} = \dfrac{\sin 70^\circ}{9}\).
\(\sin B = \dfrac{7\sin 70^\circ}{9} = 0.7308\ldots\)
\(B = \sin^{-1}(0.7308\ldots) = 46.96\ldots^\circ\). Since \(b < a\), \(B\) must be smaller than \(A\), so the acute answer is the only one.
Answer: \(B = 47.0^\circ\) (1 d.p.)
Examiner trap
Pairing a side with an angle that is NOT opposite it. Write the triangle's letters on your sketch and check: side \(a\) must face angle \(A\). Using \(\sin 73^\circ\) where the opposite angle is really \(55^\circ\) is the commonest lost mark.
Section 02
Obtuse angles and the ambiguous case.
Because \(\sin(180^\circ - x) = \sin x\), the equation \(\sin B = 0.86\) has two answers between \(0^\circ\) and \(180^\circ\): an acute one and an obtuse one. Sometimes both make a valid triangle. Try it in the ambiguous case tab of the lab: the same two sides and angle can swing into two different triangles.
Worked example Two possible triangles
In triangle \(ABC\), \(A = 35^\circ\), \(a = 6\) cm and \(b = 9\) cm. Find the two possible values of angle \(B\).
\(\sin B = \dfrac{9\sin 35^\circ}{6} = 0.8603\ldots\)
Check the obtuse one fits: \(35^\circ + 120.64^\circ = 155.64^\circ < 180^\circ\), so \(C = 24.4^\circ\) and a triangle exists.
Answer: \(B = 59.4^\circ\) or \(B = 120.6^\circ\)
Worked example When there is only one answer
In triangle \(PQR\), \(q = 8.3\) cm, \(r = 6.1\) cm and \(Q = 112^\circ\). Find angle \(R\).
\(\sin R = \dfrac{6.1\sin 112^\circ}{8.3} = 0.6814\ldots\), so \(R = 42.96\ldots^\circ\).
The obtuse option \(180^\circ - 42.96^\circ = 137.04^\circ\) is impossible: a triangle cannot have two obtuse angles.
Answer: \(R = 43.0^\circ\) (1 d.p.)
Your turn
Explain why no triangle exists with \(A = 50^\circ\), \(a = 4\) cm and \(b = 7\) cm.
\(\sin B = \dfrac{7\sin 50^\circ}{4} = 1.34\), but a sine can never be bigger than 1. Side \(a\) is too short to reach the third side.
Examiner trap
Always giving the calculator's acute angle. If the question says the angle is obtuse, or your diagram shows it is, the answer is \(180^\circ\) minus the calculator value.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Which equation is the sine rule for triangle \(ABC\)?
Quick check+10 XP
In triangle \(ABC\), \(a = 10\) cm, \(A = 30^\circ\) and \(B = 45^\circ\). Find \(b\).
Quick check+10 XP
When can you use the sine rule?
Quick check+10 XP
\(\sin 50^\circ = 0.766\). Which other angle between \(0^\circ\) and \(180^\circ\) has a sine of 0.766?
Section 03
The sine rule with bearings.
Bearings problems often give two angles and one distance, which is exactly the sine rule's job. Draw north lines at each point, work out the angles INSIDE the triangle, then apply the rule.
Worked example Locating a ship
Coastguard station \(B\) is 20 km due east of station \(A\). A ship \(S\) is on a bearing of \(040^\circ\) from \(A\) and \(300^\circ\) from \(B\). Find the distance \(AS\).
At \(A\): \(AB\) points east (\(090^\circ\)), so angle \(SAB = 90^\circ - 40^\circ = 50^\circ\).
At \(B\): \(BA\) points west (\(270^\circ\)), so angle \(SBA = 300^\circ - 270^\circ = 30^\circ\).
Angle \(ASB = 180^\circ - 50^\circ - 30^\circ = 100^\circ\). This faces \(AB = 20\) km, a complete pair.
Putting the bearing itself (\(040^\circ\) or \(300^\circ\)) into the sine rule. Bearings are measured from north; the sine rule needs the angles inside the triangle.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to IGCSE Paper 2 and Paper 4 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
In triangle \(PQR\), angle \(P = 52^\circ\), angle \(Q = 73^\circ\) and \(PQ = 10\) cm. Find \(QR\).
Exam style+20 XP
In triangle \(ABC\), \(a = 12\) cm, \(c = 15\) cm and \(A = 48^\circ\). Angle \(C\) is obtuse. Find \(C\).
Exam style+20 XP
Station \(B\) is 20 km due east of station \(A\). A ship is on a bearing of \(040^\circ\) from \(A\) and \(300^\circ\) from \(B\). How far is the ship from \(B\)?
Exam style+20 XP
In triangle \(ABC\), \(A = 30^\circ\), \(a = 5\) cm and \(b = 8\) cm. Find all possible values of angle \(B\).
Skill unlocked
The sine rule, sorted.
This skill is now recorded on your profile. Keep the chain going.