Grasp Maths

Year 11

Sine rule and cosine rule

Apply asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B} and c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C to non-right-angled triangles.

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Lesson overview

Geometry - trigonometry

In right-angled triangles, we use basic trigonometry (SOH CAH TOA). But most triangles are not right-angled. The sine rule and cosine rule allow us to find missing sides and angles in any triangle. The sine rule asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} relates sides and their opposite angles. The cosine rule, c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C, generalises Pythagoras' theorem to non-right-angled triangles. Choosing which rule to use depends on which measurements you know: use the sine rule when you have a side and its opposite angle, and the cosine rule when you have two sides and the angle between them (or all three sides).

Using the sine rule to find a missing side

In a triangle, angle A=30°A = 30°, angle B=45°B = 45°, and side a=5a = 5 cm. Find side bb.

a = 5 cmb = ?A = 30°B = 45°

The sine rule connects a side with its opposite angle. Substitute known values and solve for bb.

Using the sine rule to find a missing angle

In a triangle, side a=8a = 8 cm, side b=10b = 10 cm, and angle A=40°A = 40°. Find angle BB.

a = 8 cmb = 10 cmA = 40°B = ?

Rearrange the sine rule to solve for sin⁡B\sin B. Remember: there may be two possible angles (ambiguous case).

Using the cosine rule to find a missing side

In a triangle, sides a=7a = 7 cm and b=9b = 9 cm, and the angle between them C=50°C = 50°. Find side cc.

b = 9 cmc = ?a = 7 cmC = 50°

The cosine rule requires two sides and the included angle. Substitute and solve.

Using the cosine rule to find a missing angle

In a triangle, all three sides are a=5a = 5 cm, b=6b = 6 cm, and c=7c = 7 cm. Find angle CC.

a = 5 cmc = 7 cmb = 6 cmC = ?

Rearrange the cosine rule to solve for the angle. When you know all three sides, cosine rule is the right choice.

Worked example: Choosing and applying the right rule

A surveyor needs to find the distance across a river. She measures 60 m along one bank (line AB), then measures angle CAB = 35° and angle CBA = 80°, where C is a point on the opposite bank. How far is it across the river (distance AC)?

AB = 60 mAC = ?A = 35°B = 80°
KnownAB = 60 m, angle CAB = 35°, angle CBA = 80°

We need to find AC

First, find angle ACB180° - 35° - 80° = 65°

Sum of angles in a triangle = 180°

  1. We know two angles (35° and 80°) and the side between them (60 m, which is AB). The third angle is: angle ACB = 180° - 35° - 80° = 65°.
  2. We need side AC. This is opposite angle B (80°).
  3. Use the sine rule: ABsin⁡C=ACsin⁡B\frac{AB}{\sin C} = \frac{AC}{\sin B}.
  4. Substitute: 60sin⁡65°=ACsin⁡80°\frac{60}{\sin 65°} = \frac{AC}{\sin 80°}.
  5. Rearrange: AC=60sin⁡80°sin⁡65°AC = \frac{60 \sin 80°}{\sin 65°}.
  6. Calculate: AC=60×0.9850.906=59.10.906≈65.2AC = \frac{60 \times 0.985}{0.906} = \frac{59.1}{0.906} ≈ 65.2 m.
  7. The distance across the river is approximately 65.2 m.

Try it

Identify whether to use the sine rule or cosine rule, then solve. Remember: sine rule for (side, opposite angle) relationships; cosine rule for (two sides + included angle) or (three sides).

Question 1

In a triangle, angle A=50°A = 50°, angle B=70°B = 70°, and side a=12a = 12 cm. Which rule should you use to find side bb?

a = 12 cmb = ?A = 50°B = 70°

💡 You have two angles and a side opposite one of them. This is the perfect setup for the sine rule.

Question 2

In a triangle, sides a=8a = 8 cm, b=10b = 10 cm, and the angle between them is C=55°C = 55°. Find side cc. Which of these is correct?

b = 10 cmc = ?a = 8 cmC = 55°

💡 Use cosine rule: c2=82+102−2(8)(10)cos⁡55°=64+100−160(0.574)≈164−91.8=72.2c^2 = 8^2 + 10^2 - 2(8)(10)\cos 55° = 64 + 100 - 160(0.574) ≈ 164 - 91.8 = 72.2, so c≈8.5c ≈ 8.5 cm.

Question 3

In triangle ABC, angle A=35°A = 35°, side a=10a = 10 cm, and side b=13b = 13 cm. Find angle BB using the sine rule.

a = 10 cmb = 13 cmA = 35°B = ?

💡 sin⁡B=bsin⁡Aa=13sin⁡35°10=13×0.57410≈0.746\sin B = \frac{b \sin A}{a} = \frac{13 \sin 35°}{10} = \frac{13 \times 0.574}{10} ≈ 0.746. So B≈48.3°B ≈ 48.3° or B≈131.7°B ≈ 131.7°. The closest is the first option.

Question 4

In a triangle with sides a=5a = 5 cm, b=7b = 7 cm, and c=9c = 9 cm, find angle AA using the cosine rule.

c = 9 cma = 5 cmb = 7 cmA = ?

💡 Rearrange cosine rule: a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc\cos A, so cos⁡A=b2+c2−a22bc=49+81−252(7)(9)=105126≈0.833\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{49 + 81 - 25}{2(7)(9)} = \frac{105}{126} ≈ 0.833. So A≈33.6°A ≈ 33.6°.

Question 5

A triangle has sides 6 cm, 8 cm, and 10 cm. Is this a right-angled triangle?

6 cm8 cm10 cm

💡 Check Pythagoras: 36+64=100=10036 + 64 = 100 = 100. Yes, this is a right-angled triangle (3-4-5 triangle scaled by 2).

Common mistakes

Watch for these when working through the lesson.

  • Confusing which rule to use. Sine rule: use when you have a side and its opposite angle. Cosine rule: use when you have two sides and the included angle, or all three sides.
  • Forgetting that the sine rule can produce two possible angles (the ambiguous case). If sin⁡B=0.5\sin B = 0.5, then B≈30°B ≈ 30° or B≈150°B ≈ 150°. Check which is valid for your triangle.
  • Not checking units and rounding appropriately. Always work in the same unit (e.g., cm or m) and round your final answer to a reasonable accuracy (often to 1 d.p. for lengths, or to nearest degree for angles).

Related topics

These ideas fit closely with this lesson.

  • Trigonometry in right-angled triangles
  • Area of non-right-angled triangles using $\frac{1}{2}ab\sin C$
  • 3D trigonometry and angle between planes

Practice next

Independent practice will plug in here

This lesson builds the understanding first. Deeper adaptive practice can sit here later.