CourseCourseQ BankQuestion BankDocsDocuments

E6.5 Non-right-angled triangles

Syllabus
0580–2028–2029
Topic
E6.5
Level
Extended

Choose the sine rule or cosine rule

The sine and cosine rules extend trigonometry to triangles without a right angle. Method choice depends on which sides and angles are known—not on the triangle's orientation.

Known information Rule and useful form
a side and its opposite angle, plus another side or angle sine rule: asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}
two sides and their included angle; find the third side cosine rule: c2=a2+b2−2abcos⁡Cc^2=a^2+b^2-2ab\cos C
all three sides; find an angle cosine rule: cos⁡C=a2+b2−c22ab\cos C=\dfrac{a^2+b^2-c^2}{2ab}

Label each side opposite its matching capital angle. For the sine rule, use two complete opposite pairs and rearrange. For the cosine rule, make the required side or angle the c,Cc,C pair; CC must be the angle between sides aa and bb. Keep unrounded values until the final answer.

With sides 6.46.4 cm and 10.910.9 cm enclosing 38∘38^\circ, the third side is c=6.42+10.92−2(6.4)(10.9)cos⁡38∘=7.06c=\sqrt{6.4^2+10.9^2-2(6.4)(10.9)\cos38^\circ}=7.06 cm. If instead A=50∘A=50^\circ, B=100∘B=100^\circ and b=12b=12 cm, then a=12sin⁡50∘/sin⁡100∘=9.33a=12\sin50^\circ/\sin100^\circ=9.33 cm.

The longest side must face the largest angle, which is a useful check. When inverse sine gives an angle, its supplement has the same sine; use the stated geometry and angle sum to decide whether an acute or obtuse value is valid. Cosine rule resolves an SSS angle directly.

Use the sine area formula and handle ambiguity

Two sides and their included angle determine a triangle's area because one side contributes the perpendicular height bsin⁡Cb\sin C to the other side used as the base.

K=\frac12ab\sin C

Choose two known sides aa and bb and use the angle CC between them. Substitute consistently and attach square units. For a reverse problem, rearrange to sin⁡C=2K/(ab)\sin C=2K/(ab), find the acute calculator angle, then check its supplement 180∘−C180^\circ-C because both angles have the same sine.

For sides 88 cm and 99 cm with included angle 50∘50^\circ, K=12(8)(9)sin⁡50∘=27.6K=\frac12(8)(9)\sin50^\circ=27.6 cm2^2. If sides 1010 cm and 1414 cm enclose an unknown angle and K=45K=45 cm2^2, then sin⁡C=90/140\sin C=90/140. This gives C=40.0∘C=40.0^\circ or 140.0∘140.0^\circ: an acute/obtuse ambiguous pair.

The angle must be included between the two sides used in the formula. Since sin⁡C=sin⁡(180∘−C)\sin C=\sin(180^\circ-C), area alone may not determine a unique angle. Reject any candidate that conflicts with other given lengths, angles or the triangle angle sum.