Sine Rule and Cosine Rule Examples: Surveying, Bearings and the Ambiguous Case

In 30 seconds

The sine rule and cosine rule solve triangles that have no right angle. Label each side with the same letter as the angle opposite it: side $ a $ faces angle $ A $.
Use the cosine rule, $ a^2 = b^2 + c^2-2bc\cos A $, when you know two sides and the angle between them, or all three sides.
Use the sine rule, $ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} $, when you know a side and the angle opposite it, plus one more side or angle. The area is $ \tfrac{1}{2}ab\sin C $.
Case: a triangular plot with sides 120 m and 85 m meeting at 64° has a third side of 112.6 m and an area of about 4,584 m².
Watch for the ambiguous case: finding an angle from its sine can give two possible triangles.

The case: surveying a triangular plot

A family has inherited a triangular plot of land on the edge of a village and wants to fence it and work out its area before selling. A surveyor stands at corner $ A $, where two straight boundaries meet. With a tape and a theodolite she measures:

the boundary $ AB = 120 $ m,
the boundary $ AC = 85 $ m,
the angle between them, $ \angle BAC = 64^\circ $.

The third boundary, $ BC $, runs across a stream and through thick hedge, so it cannot be taped directly. The family’s questions: how long is the third side, what are the other two angles, what is the area of the plot, and roughly what will it cost to fence all the way round at $18 per metre? The plot and the prices are illustrative.

The triangle has no right angle, so Pythagoras and SOHCAHTOA cannot be used directly. This is exactly what the sine and cosine rules are for.

The concept: trigonometry for any triangle

A triangle is fixed once you know enough about it. Two sides and the angle between them pin it down completely: there is only one way to join the free ends. Three sides also fix it. So does one side and two angles. In each case, the rest of the triangle can be calculated. The sine and cosine rules are the tools that do that calculation.^[1]

The cosine rule is Pythagoras with a correction term. If angle $ A $ is $ 90^\circ $, then $ \cos A = 0 $ and the rule becomes $ a^2 = b^2 + c^2 $. If $ A $ is smaller than $ 90^\circ $, the side opposite is shorter than Pythagoras would say, and the term $ -2bc\cos A $ takes some length off. If $ A $ is obtuse, $ \cos A $ is negative and the side gets longer.

The sine rule says that in any triangle, each side divided by the sine of its opposite angle gives the same number. Bigger angles face longer sides, and the sine rule makes that exact.

Key termSolving a triangle: finding all the unknown sides and angles of a triangle from the ones you know. The labelling convention is that side $ a $ is opposite angle $ A $, side $ b $ is opposite angle $ B $, and side $ c $ is opposite angle $ C $.

The formulas

\[ a^2 = b^2 + c^2-2bc\cos A \] \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] \[ \text{Area} = \tfrac{1}{2}ab\sin C \]

Here $ a $, $ b $ and $ c $ are side lengths and $ A $, $ B $ and $ C $ are the angles opposite them. The cosine rule can be rearranged to find an angle from three sides: $ \cos A = \dfrac{b^2 + c^2-a^2}{2bc} $. The area formula uses two sides and the angle between them, whichever pair that is.^[2]

Chapter 7 of Mathematics Made Simple covers the sine rule, cosine rule and triangle area with clear diagrams.Get the textbook · $12

Why these formulas, and which one first

The choice depends on what you know, not on what you want:

You know	Start with
Two sides and the angle between them (SAS)	Cosine rule, to find the third side
All three sides (SSS)	Cosine rule, rearranged, to find an angle
Two angles and any side (AAS or ASA)	Angle sum $ 180^\circ $, then the sine rule
Two sides and an angle not between them (SSA)	Sine rule, and check for the ambiguous case

The surveyor measured two sides and the angle between them, so the case is SAS. The sine rule cannot start the job, because every pair in it needs a side and its opposite angle, and we do not have any such pair yet. The cosine rule needs exactly what we have.

Neither rule is needed if the triangle has a right angle: SOHCAHTOA is quicker. And if you only want the area and you already know two sides and the angle between them, go straight to $ \tfrac{1}{2}ab\sin C $.

Setting it up

Label the triangle so the known angle is $ A $. Then the side opposite it, the unknown boundary $ BC $, is $ a $. The other two sides are $ b = AC = 85 $ m (opposite $ B $) and $ c = AB = 120 $ m (opposite $ C $). Figure 1 shows the plot to scale.

The plot to scale, with the hidden boundary BC metres

Figure 1. The surveyor measured AB, AC and the 64° angle between them (solid). The cosine rule gives the hidden boundary BC (dashed), the sine rule gives the other angles, and ½bc sin A gives the area.

Two practical points before calculating. Make sure your calculator is in degree mode, since the angle was measured in degrees. And keep the full calculator value of $ a $ for later steps; rounding it to 112.6 too soon can shift the angles in the last decimal place.

Solving it

Worked example

In triangle $ ABC $, $ b = 85 $ m, $ c = 120 $ m and $ A = 64^\circ $. Find $ a $, angles $ B $ and $ C $, the area, and the cost of fencing the perimeter at $18 per metre.

Step 1Cosine rule: $ a^2 = 85^2 + 120^2 $ $ {}- 2 \times 85 \times 120 \times \cos 64^\circ $.

Step 2$ 85^2 + 120^2 = 7225 + 14400 = 21625 $, and $ 20400\cos 64^\circ \approx 8942.77 $.

Step 3$ a^2 \approx 21625-8942.77 = 12682.23 $, so $ a \approx 112.6 $ m.

Step 4Sine rule for the smaller unknown angle, $ B $, opposite the shortest side: $ \sin B = \dfrac{85\sin 64^\circ}{112.615} \approx 0.6784 $.

Step 5$ B \approx 42.7^\circ $. It must be acute, because it faces the shortest side.

Step 6Angle sum: $ C = 180^\circ-64^\circ-42.7^\circ = 73.3^\circ $.

Step 7Area: $ \tfrac{1}{2} \times 85 \times 120 \times \sin 64^\circ $ $ {}= 5100\sin 64^\circ \approx 4584 $ m².

Step 8Perimeter: $ 112.6 + 85 + 120 = 317.6 $ m. Fencing: $ 317.6 \times 18 \approx \$5717 $.

The hidden boundary is about 112.6 m long, and the plot covers about 4,584 m², or just under half a hectare.

The same two rules are the workhorses of navigation. Bearings are measured clockwise from north, as three-figure angles, and the main skill is turning the bearings into an angle inside the triangle.

Worked example

A boat leaves harbour $ H $ and sails 8 km on a bearing of $ 030^\circ $ to a buoy $ P $. It then sails 11 km on a bearing of $ 115^\circ $ to an island $ Q $. How far is the island from the harbour, and on what bearing?

Step 1Angle at $ P $: the bearing from $ P $ back to $ H $ is $ 030^\circ + 180^\circ = 210^\circ $. The boat leaves $ P $ on $ 115^\circ $. So $ \angle HPQ = 210^\circ-115^\circ = 95^\circ $.

Step 2Cosine rule: $ HQ^2 = 8^2 + 11^2 $ $ {}- 2 \times 8 \times 11 \times \cos 95^\circ $.

Step 3$ \cos 95^\circ $ is negative, so the last term adds: $ HQ^2 \approx 185 + 15.34 = 200.34 $, giving $ HQ \approx 14.15 $ km.

Step 4Sine rule for the angle at $ H $: $ \sin \angle PHQ = \dfrac{11\sin 95^\circ}{14.154} \approx 0.7742 $, so $ \angle PHQ \approx 50.7^\circ $.

Step 5Bearing of $ Q $ from $ H $: $ 030^\circ + 50.7^\circ = 080.7^\circ $.

The island is about 14.15 km from the harbour on a bearing of about $ 080.7^\circ $. To sail straight back, the boat would steer $ 080.7^\circ + 180^\circ = 260.7^\circ $.

Step 4 was safe: the angle at $ H $ faces the 11 km side, which is not the longest side, so it cannot be obtuse. That check matters, because the sine rule has a trap.

Worked example

Exam style: the ambiguous case. In triangle $ ABC $, $ a = 7 $ cm, $ b = 10 $ cm and $ A = 35^\circ $. Find the possible values of angle $ B $ and of side $ c $.

Step 1Sine rule: $ \sin B = \dfrac{10\sin 35^\circ}{7} \approx 0.8194 $.

Step 2The calculator gives $ B \approx 55.0^\circ $. But $ \sin(180^\circ-B) = \sin B $, so $ B \approx 125.0^\circ $ is also possible.

Step 3Test both: $ 35^\circ + 125.0^\circ < 180^\circ $, so both triangles exist.

Step 4If $ B \approx 55.0^\circ $: $ C \approx 90.0^\circ $ and $ c = \dfrac{7\sin C}{\sin 35^\circ} \approx 12.20 $ cm.

Step 5If $ B \approx 125.0^\circ $: $ C \approx 20.0^\circ $ and $ c = \dfrac{7\sin C}{\sin 35^\circ} \approx 4.18 $ cm.

Two different triangles fit the same data. The side of length 7 can swing to meet the third side in two places (Figure 2).

Ambiguous case: a = 7, b = 10, A = 35° cm

Figure 2. Side a = 7 swings about C (grey circle) and meets the 35° line from A twice. Both points give a valid triangle, so the sine rule has two answers for B.

Master bearings and the ambiguous case with 1,565 practice questions, each fully worked.Get 1,565 questions · $9

What the answer tells us

For the family, the results are practical. The third boundary is about 112.6 m, so the plot’s perimeter is about 317.6 m and fencing will cost roughly $5,700 at $18 a metre. The area, about 4,584 m², is the number a buyer or a land registry will care about, and it is just under half a hectare (5,000 m²).

Several checks say the answers are reasonable. The angles add to $ 64^\circ + 42.7^\circ + 73.3^\circ = 180^\circ $. The largest angle, $ C $, faces the longest side, $ c = 120 $ m, and the smallest, $ B $, faces the shortest, $ b = 85 $ m. And $ BC $ lies between $ 120-85 = 35 $ m and $ 120 + 85 = 205 $ m, as the triangle inequality requires.

How sensitive is the area to a small measuring error? The area is $ 5100\sin A $. If the angle were really $ 65^\circ $ instead of $ 64^\circ $, the area would be $ 5100\sin 65^\circ \approx 4622 $ m², about 38 m² more. A one-degree error changes the area by under 1%, which is reassuring for a sale, but it is still worth measuring the angle twice.

Common mistakeUsing the sine rule to find an obtuse angle without thinking. A calculator’s $ \sin^{-1} $ only ever returns an angle between $ 0^\circ $ and $ 90^\circ $. If the angle you want could be obtuse, either use the cosine rule (whose $ \cos^{-1} $ gives the right angle directly) or find the smallest angle first with the sine rule, since it can never be obtuse.

Common mistakeMislabelling the sides. In $ a^2 = b^2 + c^2-2bc\cos A $, the side on the left must be the one opposite the angle in the cosine. Write the letters on your sketch before you substitute anything, and use degree mode throughout.

Exam tipIn IB Mathematics: Analysis and Approaches, the sine rule (including the ambiguous case), the cosine rule and the area $ \tfrac{1}{2}ab\sin C $ are SL topic 3.2, and applications with bearings and angles of elevation are SL 3.3.^[3] In Cambridge 9709, the rules are treated as knowledge you bring from earlier courses rather than a separate Paper 1 topic, and they are used freely inside trigonometry and mechanics questions.^[4] They are not part of AP Calculus. In any paper, a clear labelled sketch usually earns method marks even if the arithmetic slips.

For more on triangles, radians and the unit circle, see Chapter 7 of the Mathematics Made Simple textbook. If you want plenty of bearings, area and ambiguous-case practice, the Mathematics Made Simple practice book has graded questions with every step shown.

Get both books in the Mathematics Made Simple bundle for $16 and save $5 on trigonometry revision.Both books · $16

Practice questions

EasyQ1. In triangle $ ABC $, $ b = 6 $ cm, $ c = 9 $ cm and $ A = 40^\circ $. Find $ a $ to 3 significant figures. Show worked answer

Cosine rule: $ a^2 = 6^2 + 9^2 $ $ {}- 2 \times 6 \times 9 \times \cos 40^\circ $.
$ a^2 = 117-108\cos 40^\circ \approx 117-82.73 = 34.27 $.
$ a \approx 5.85 $ cm.

EasyQ2. In triangle $ ABC $, $ A = 40^\circ $, $ B = 75^\circ $ and $ a = 12 $ cm. Find $ b $ to 3 significant figures. Show worked answer

Sine rule: $ \dfrac{b}{\sin 75^\circ} = \dfrac{12}{\sin 40^\circ} $.
$ b = \dfrac{12\sin 75^\circ}{\sin 40^\circ} \approx 18.0 $ cm.

MediumQ3. A triangle has sides 5 cm, 7 cm and 8 cm. Find the angle opposite the 7 cm side, and the exact area of the triangle. Show worked answer

$ \cos\theta = \dfrac{5^2 + 8^2-7^2}{2 \times 5 \times 8} = \dfrac{40}{80} = \dfrac{1}{2} $, so $ \theta = 60^\circ $.
Area $ = \tfrac{1}{2} \times 5 \times 8 \times \sin 60^\circ $
$ {}= 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32 $ cm².

MediumQ4. In triangle $ ABC $, $ a = 6 $ cm, $ b = 8 $ cm and $ A = 40^\circ $. Show that there are two possible triangles and find both values of $ c $. Show worked answer

$ \sin B = \dfrac{8\sin 40^\circ}{6} \approx 0.8571 $, so $ B \approx 59.0^\circ $ or $ B \approx 121.0^\circ $.
$ 40^\circ + 121.0^\circ < 180^\circ $, so both are possible.
$ B \approx 59.0^\circ $: $ C \approx 81.0^\circ $, $ c = \dfrac{6\sin C}{\sin 40^\circ} \approx 9.22 $ cm.
$ B \approx 121.0^\circ $: $ C \approx 19.0^\circ $, $ c \approx 3.04 $ cm.

ExamQ5. A ship sails from port $ P $ for 20 km on a bearing of $ 050^\circ $ to a point $ Q $, then 15 km on a bearing of $ 170^\circ $ to a point $ R $. (a) Show that $ \angle PQR = 60^\circ $. [2] (b) Find the exact distance $ PR $, and give it to 3 significant figures. [2] (c) Find the bearing of $ R $ from $ P $. [3] (d) Find the area of triangle $ PQR $. [1] Show worked answer

(a) The bearing of $ P $ from $ Q $ is $ 050^\circ + 180^\circ = 230^\circ $ [1]. So $ \angle PQR = 230^\circ-170^\circ = 60^\circ $ [1].
(b) $ PR^2 = 20^2 + 15^2-2 \times 20 \times 15 \times \cos 60^\circ $
$ {}= 625-300 = 325 $ [1], so $ PR = \sqrt{325} = 5\sqrt{13} \approx 18.0 $ km [1].
(c) $ \sin\angle QPR = \dfrac{15\sin 60^\circ}{\sqrt{325}} $ [1], so $ \angle QPR \approx 46.1^\circ $, acute because it faces the 15 km side, which is not the longest [1]. Bearing $ = 050^\circ + 46.1^\circ = 096.1^\circ $ [1].
(d) Area $ = \tfrac{1}{2} \times 20 \times 15 \times \sin 60^\circ = 75\sqrt{3} \approx 129.9 $ km² [1].

Frequently asked questions

How do I know whether to use the sine rule or the cosine rule?

Look for a “pair”: a side together with the angle opposite it. If you know a complete pair, the sine rule will work. If you do not (two sides and the angle between them, or three sides), start with the cosine rule.

What is the ambiguous case of the sine rule?

When you know two sides and an angle that is not between them, there may be two different triangles that fit. It happens because $ \sin B = \sin(180^\circ-B) $. Always test whether $ 180^\circ-B $ also leaves room for the third angle.

Does the cosine rule work for right-angled triangles?

Yes. If $ A = 90^\circ $, $ \cos A = 0 $ and the cosine rule becomes $ a^2 = b^2 + c^2 $, which is Pythagoras’ theorem. The cosine rule is the general version.

Can I find the area of a triangle from three sides?

Yes. Use the cosine rule to find one angle, then $ \tfrac{1}{2}ab\sin C $ with the two sides that enclose it. Question 3 above does exactly this.

References

Abramson, J. (2017). Precalculus. Houston, TX: OpenStax. Chapter: Further Applications of Trigonometry (Non-right Triangles: Law of Sines; Law of Cosines). https://openstax.org/details/books/precalculus
Stewart, J., Redlin, L. and Watson, S. (2016). Precalculus: Mathematics for Calculus, 7th edition. Boston, MA: Cengage Learning.
International Baccalaureate Organization (2019). Mathematics: analysis and approaches guide (first assessment 2021). Cardiff: IBO.
Cambridge International. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. Cambridge: Cambridge University Press & Assessment.

Sine Rule and Cosine Rule Examples: Surveying, Bearings and the Ambiguous Case

The case: surveying a triangular plot

The concept: trigonometry for any triangle

The formulas

Why these formulas, and which one first

Setting it up

Solving it

What the answer tells us

Practice questions

Frequently asked questions

How do I know whether to use the sine rule or the cosine rule?

What is the ambiguous case of the sine rule?

Does the cosine rule work for right-angled triangles?

Can I find the area of a triangle from three sides?

References

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