A-level Maths revision · Pure, year 1

Sine rule, cosine rule and area

The idea

For any triangle: sine rule asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B}; cosine rule a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc\cos A; area =12absin⁡C= \frac{1}{2}ab\sin C. When the sine rule finds an angle, 180∘180^\circ minus that angle may also fit.

  1. Two sides and the angle between them, or three sides: use the cosine rule.
  2. A side and its opposite angle known: use the sine rule.
  3. Check the ambiguous case: does 180∘−B180^\circ - B still leave room in the triangle?

Watch out The sine rule can give two possible triangles; check the obtuse option.

See it worked

Worked example

In triangle ABCABC, a=7a = 7, b=9b = 9 and A=40∘A = 40^\circ. Find the possible values of BB.

  1. sin⁡B9=sin⁡40∘7\frac{\sin B}{9} = \frac{\sin 40^\circ}{7}, so sin⁡B=0.8264\sin B = 0.8264.
  2. B=55.7∘B = 55.7^\circ, or 180∘−55.7∘=124.3∘180^\circ - 55.7^\circ = 124.3^\circ.
  3. Both leave a positive third angle, so both are possible.

Answer B=55.7∘B = 55.7^\circ or 124.3∘124.3^\circ

Your turn

Question 1Non-calculator3 marks

In triangle ABCABC, b=5b = 5, c=8c = 8 and A=60∘A = 60^\circ. Find aa.

Question 2Non-calculator3 marks

Find the area of a triangle with sides 6 cm6\text{ cm} and 9 cm9\text{ cm} and an angle of 30∘30^\circ between them.

Question 3Calculator allowed3 marks

A triangle has sides 44, 55 and 66. Find its largest angle to 1 decimal place.

In past papers

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