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Learn4024Paper 1 · Non-calculatorTrigonometrySine rule, cosine rule and area
Trigonometry · Subtopic 3 of 4

Sine Rule, Cosine Rule and Area

For triangles without a right angle: use the sine rule when you know a side and its opposite angle, and the cosine rule when you know two sides and the angle between them, or all three sides. The area is 12absin⁡C\tfrac12ab\sin C.

Cambridge 4024 Paper 1Cambridge 4024 Paper 2Edexcel 4MA1 Papers 1H, 2HAQA 8300 Papers 1–3≈ 40 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use the sine rule to find a side or an angle.
  • Use the cosine rule to find a side or an angle.
  • Use 12absin⁡C\tfrac12ab\sin C for the area of any triangle.
  • Recognise the ambiguous case and find obtuse angles.
Pause and think

In the cosine rule a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A, what happens when A=90∘A=90^\circ?

Section 2 of 14

Core idea

Start hereTriangles without right angles

SOH CAH TOA only works with a right angle. A surveyor's triangle of fields rarely has one. The sine rule and cosine rule extend trigonometry to every triangle.

The big idea

asin⁡A=bsin⁡B=csin⁡C\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}; a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A; area 12absin⁡C\frac12ab\sin C. sin⁡(180∘−x)=sin⁡x\sin(180^\circ-x)=\sin x, cos⁡(180∘−x)=−cos⁡x\cos(180^\circ-x)=-\cos x.

1. The rules

A triangle labelled for the rules
ABCabc
Each side faces the angle with the same letter.
Area and the sine rule
  1. h=bsin⁡Ch=b\sin C
    Drop a perpendicular from AA to side aa.
  2. area=12absin⁡C\text{area}=\tfrac12ab\sin C
    Half base × height.
  3. 12bcsin⁡A=12casin⁡B=12absin⁡C\tfrac12bc\sin A=\tfrac12ca\sin B=\tfrac12ab\sin C
    Same area from any side.
  4. asin⁡A=bsin⁡B=csin⁡C\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}
    Divide by 12abc\frac12abc, invert.
Worked exampleSine rule

A=50∘A=50^\circ, B=60∘B=60^\circ, a=8a=8.

Goal
bb and CC.
Strategy
Pair aa with AA.
  1. bsin⁡60∘=8sin⁡50∘\frac b{\sin60^\circ}=\frac8{\sin50^\circ}
    Sine rule.
  2. b=8sin⁡60∘sin⁡50∘=9.04b=\frac{8\sin60^\circ}{\sin50^\circ}=9.04
    Rearrange.
  3. C=180−50−60=70∘C=180-50-60=70^\circ
    Angle sum.
Answerb=9.04b=9.04, C=70∘C=70^\circ.
✓ CheckB>AB>A so b>ab>a ✓.
Worked exampleCosine rule with an obtuse angle

b=5b=5, c=7c=7, A=120∘A=120^\circ.

Goal
aa and the area.
Strategy
Two sides and the included angle.
  1. a2=25+49−70cos⁡120∘a^2=25+49-70\cos120^\circ
    Cosine rule.
  2. =74+35=109⇒a=10.4=74+35=109\Rightarrow a=10.4
    cos⁡120∘=−12\cos120^\circ=-\frac12: minus a negative.
  3. 12(5)(7)sin⁡120∘=15.2\tfrac12(5)(7)\sin120^\circ=15.2
    Area.
Answera=10.4a=10.4; area 15.215.2.
✓ CheckObtuse angle: opposite side longest ✓.
Common misconceptionIgnoring the sign of cos cos⁡120∘\cos120^\circ is negative, so the third side gets longer.
Pause and think

Angle opposite 7 in a triangle with sides 3, 5, 7?

Summary

  • Sine rule for a matching pair.
  • Cosine rule for SAS and SSS.
Key words
Obtuse
Between 90° and 180°.
HistoryFrom Persia. The sine rule for plane triangles was set out clearly by the Persian scholar Nasir al-Din al-Tusi in the 13th century, in one of the first books to treat trigonometry as a subject in its own right rather than as a tool for astronomy.
Fun factMeasuring Everest with triangles. The Great Trigonometrical Survey of India spent decades covering the country with a chain of triangles. In 1852 its calculations showed that "Peak XV" was the highest mountain in the world. It was later named after George Everest, the survey's former head.
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