BETAThis platform is under active development; bugs, missing features, and risk of data loss are present. Thank you for your support!

Sine Cosine Rule And Area Of Triangles

Learn it by playing

Answer these questions to earn energy, then fish and explore. No account needed.

For teachers: ready-to-use lesson slides, revision notes, diagrams for Sine Cosine Rule And Area Of Triangles (Maths [CIE], Extended) — use them in your lesson, or run the topic as a live class game.

Notes

The Sine Rule

  • Used in **non-right-angled triangles** to find missing side lengths or angles.
  • States: **a /sinA=b/sinB=c/sin/ \sin A = b / \sin B = c / \sin C**, where a is opposite A, etc.
  • To find a missing length, use two equal parts of the rule and solve.
  • To find a missing angle, rearrange to **sin A/a=sinB/b=sinC/A / a = \sin B / b = \sin C / c**.
  • **Ambiguous case**: given two sides and a non-included angle, there may be two possible triangles (acute and obtuse).
  • If the required angle is obtuse, use: **obtuse angle =180= 180^{\circ} – acute angle** from calculator.

The Cosine Rule

  • Used in **non-right-angled triangles** when you have two sides and the included angle (to find the third side) or all three sides (to find an angle).
  • For a side: **a² =b2+c2= b^{2} + c^{2} – 2bc cos A**, where A is the angle between b and c.
  • For an angle: **cos A=(b2+c2A = (b^{2} + c^{2}a2)/a^{2}) / (2bc)**.
  • No ambiguous case – the cosine rule gives a unique angle.
  • Label sides carefully: side a is opposite angle A.

Area of a Triangle

  • For any triangle: **Area = ½ ab sin C**, where C is the angle between sides a and b.
  • If C=90,sin90=1C = 90^{\circ}, \sin 90^{\circ} = 1, so Area = ½ × base × height (right-angled triangle).
  • Ensure all lengths are in the same units before calculating area.
  • If the included angle is not given, use sine or cosine rule first to find it.

Deciding the Trig Rule

  • **Sine rule**: use when you have an opposite pair (side and angle) and need another side or angle.
  • **Cosine rule**: use when you have two sides and the included angle (to find the third side) or all three sides (to find an angle).
  • **Area rule**: use when you have two sides and the included angle (to find area).
  • If no rule fits directly, use **angles in a triangle sum to 180°** to find a missing angle.
  • Harder questions may require **multiple trig rules** in sequence.

Worked Example – Sine Rule

  • Given triangle ABC with AB=8.1AB = 8.1 cm, BC=12.3BC = 12.3 cm, angle BCA=27BCA = 27^{\circ}.
  • Find angle x (at A): use sinx/12.3=sin27/8.1\sin x / 12.3 = \sin 27^{\circ} / 8.1x=x = sin⁻¹(12.3 sin 27° / 8.1) ≈ 43.6°.
  • Find side y (AC): first find angle ABC=180ABC = 180^{\circ} – 27° – 43.6=109.443.6^{\circ} = 109.4^{\circ}, then y/sin109.4=8.1/sin27y / \sin 109.4^{\circ} = 8.1 / \sin 27^{\circ} → y ≈ 16.8 cm.

Worked Example – Cosine Rule

  • Given triangle ABC with AB=4.2AB = 4.2 km, BC=3.8BC = 3.8 km, AC=7.1AC = 7.1 km.
  • Find angle ABC: use cos θ =(4.22+3.82= (4.2^{2} + 3.8^{2}7.12)/(2×4.2×3.8)7.1^{2}) / (2 \times 4.2 \times 3.8) → θ = cos⁻¹(...) ≈ 125.0°.

Worked Example – Area of a Triangle

  • Given triangle ABC with AB=32AB = 32 cm, AC=1.1mAC = 1.1 m, angle BAC=74BAC = 74^{\circ}.
  • Convert to same units: AB=0.32m,AC=1.1mAB = 0.32 m, AC = 1.1 m.
  • Area = ½ ×1.1×0.32×sin74\times 1.1 \times 0.32 \times \sin 74^{\circ}0.169m2(3s.f.)0.169 m^{2} (3 s.f.).

Worked Example – Multiple Rules

  • Find area of triangle with sides 4.4 cm, 7.4 cm, 4.8 cm.
  • Use cosine rule to find an angle: cosABC=(4.42\cos ABC = (4.4^{2}7.427.4^{2}4.82)/4.8^{2}) / (–2 ×7.4×4.8)\times 7.4 \times 4.8) → ABC ≈ 34.65°.
  • Then area = ½ ×4.8×7.4×sin34.65\times 4.8 \times 7.4 \times \sin 34.65^{\circ}10.1cm2(3s.f.)10.1 cm^{2} (3 s.f.).

Sine Rule – Triangle Labelling

ABCacbABC

Cosine Rule – Triangle Labelling

ABCacbABC

Area of a Triangle – Formula

ABCacbCArea = ½ ab sin C

Flowchart – Choosing the Rule

Start: What do you know?Two sides & included angle?Use Cosine Rule or Area RuleOpposite pair (side & angle)?Use Sine Rule

Practice questions

Free preview — 8 of 36 questions. Sign up to see them all.

  1. 1.What is the formula for the area of a triangle given two sides and the included angle?

    Easy
    • AArea =12= \frac{1}{2} ab sin C
    • BArea =12= \frac{1}{2} ab cos C
    • CArea =absinC= ab \sin C
    • DArea =12= \frac{1}{2} ab tan C
  2. 2.The sine rule states that for any triangle ABC:

    Easy
    • Aa/sinA=b/sinB=c/sinCa/\sin A = b/\sin B = c/\sin C
    • Ba/sinB=b/sinC=c/sinAa/\sin B = b/\sin C = c/\sin A
    • CsinAa=sinBb=sinCc\sin \frac{A}{a} = \sin \frac{B}{b} = \sin \frac{C}{c}
    • DasinA=bsinB=csinCa \sin A = b \sin B = c \sin C
  3. 3.Which rule should be used to find a side when you know two sides and the included angle?

    Easy
    • ACosine rule
    • BSine rule
    • CArea rule
    • DPythagoras' theorem
  4. 4.In the cosine rule a2=b2+c22bccosAa^{2} = b^{2} + c^{2} - 2bc \cos A, what does angle A represent?

    Easy
    • AThe angle opposite side a
    • BThe angle between sides b and c
    • CThe angle opposite side b
    • DThe angle between sides a and c
  5. 5.Triangle ABC has AB=8AB = 8 cm, AC=5AC = 5 cm and angle BAC=30BAC = 30^{\circ}. Find the area of the triangle.

    Medium
    • A10cm210 cm^{2}
    • B20cm220 cm^{2}
    • C40cm240 cm^{2}
    • D34.6cm234.6 cm^{2}
  6. 6.In triangle XYZ,XY=7XYZ, XY = 7 cm, XZ=9XZ = 9 cm and angle YXZ=60YXZ = 60^{\circ}. Use the cosine rule to find YZ.

    Medium
    • A49+812×7×9×cos60=13063=67\sqrt{49 + 81 - 2\times 7\times 9\times \cos 60^{\circ}} = \sqrt{130 - 63} = \sqrt{67} ≈ 8.19 cm
    • B49+812×7×9×sin60=130108.9=21.1\sqrt{49 + 81 - 2\times 7\times 9\times \sin 60^{\circ}} = \sqrt{130 - 108.9} = \sqrt{21.1} ≈ 4.59 cm
    • C49+812×7×9×cos30=130108.9=21.1\sqrt{49 + 81 - 2\times 7\times 9\times \cos 30^{\circ}} = \sqrt{130 - 108.9} = \sqrt{21.1} ≈ 4.59 cm
    • D49+812×7×9×tan60=130218.2=\sqrt{49 + 81 - 2\times 7\times 9\times \tan 60^{\circ}} = \sqrt{130 - 218.2} = negative
  7. 7.In triangle PQR,PQ=10PQR, PQ = 10 cm, PR=14PR = 14 cm and angle PQR=40PQR = 40^{\circ}. Which rule can be used to find angle PRQ?

    Medium
    • ASine rule
    • BCosine rule
    • CArea rule
    • DPythagoras' theorem
  8. 8.A triangle has sides of lengths 5 cm, 6 cm and 7 cm. Find the angle opposite the side of length 7 cm.

    Medium
    • A78.5°
    • B44.4°
    • C57.1°
    • D101.5°

Unlock all 36 questions, slides & more

Create a free account to see every question, the slides, flashcards and revision notes for this topic.

Past papers

Past-paper practice for this topic is coming soon.

🗂️ Coming soon