TMUA: Sine rule, cosine rule and area
The sine rule, the cosine rule and the area of a triangle.
- Section 1 Part 1 · MM4 Trigonometry · spec MM4.1
- Papers 1 and 2
- 8 practice questions
- No calculator
What the specification covers
- Sine and cosine rules, ½ab sin C
- The ambiguous case (angle-side-side)
- 2D and 3D problems
Key ideas
- Sine rule: \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\). Cosine rule: \(c^2=a^2+b^2-2ab\cos C\).
- Area \(=\tfrac12ab\sin C\).
- Ambiguous case: given an angle, the side next to it and the side opposite it, there can be 0, 1 or 2 triangles.
- Know \(\cos 60^\circ=\tfrac12\), \(\cos120^\circ=-\tfrac12\), \(\sin30^\circ=\tfrac12\).
Common mistakes
- With \(C=120^\circ\), \(-2ab\cos C=+ab\): the sign flips.
- In the ambiguous case, check both angles \(B\) and \(180^\circ-B\) against the angle sum.
Exam tip
Label the triangle before substituting, so the angle is opposite the side you want.
Worked example
Worked example
In triangle \(ABC\), \(BC=7\), \(CA=8\) and angle \(ACB=120^\circ\). What is the length \(AB\)?
- \(\sqrt{113}\)
- \(\sqrt{57}\)
- \(13\)
- \(12\)
- \(15\)
Answer: C: \(13\)
Cosine rule: \(AB^2=49+64-2\cdot7\cdot8\cos120^\circ=113+56=169\), so \(AB=13\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
In triangle \(ABC\), angle \(BAC=30^\circ\), \(AB=10\) and \(BC=x\). For which values of \(x\) are there exactly two non-congruent triangles \(ABC\)?
- \(5 \le x \le 10\)
- \(x > 10\)
- \(5 < x < 10\)
- \(0 < x < 5\)
- \(x > 5\)
Show the answer and solution
Answer: C: \(5 < x < 10\)
Put \(A\) at the origin with \(AC\) along a ray at \(30^\circ\) to \(AB\). The distance from \(B\) to that ray is \(10\sin30^\circ=5\). The circle centre \(B\), radius \(x\), meets the ray at two points on the ray (beyond \(A\)) exactly when \(5
Question 2
In triangle \(ABC\), \(BC=8\), \(CA=3\) and angle \(ACB=60^\circ\). What is the length \(AB\)?
- \(10\)
- \(\sqrt{73}\)
- \(7\)
- \(5\)
- \(\sqrt{97}\)
Show the answer and solution
Answer: C: \(7\)
Cosine rule: \(AB^2=8^2+3^2-2\cdot8\cdot3\cos60^\circ=73-24=49\), so \(AB=7\).
Question 3
In triangle \(ABC\), \(BC=15\), \(CA=7\) and angle \(ACB=60^\circ\). What is the length \(AB\)?
- \(13\)
- \(8\)
- \(\sqrt{274}\)
- \(\sqrt{379}\)
- \(21\)
Show the answer and solution
Answer: A: \(13\)
Cosine rule: \(AB^2=15^2+7^2-2\cdot15\cdot7\cos60^\circ=274-105=169\), so \(AB=13\).
Question 4
In triangle \(ABC\), \(BC=6\), \(CA=10\) and angle \(ACB=120^\circ\). What is the length \(AB\)?
- \(2 \sqrt{19}\)
- \(2 \sqrt{34}\)
- \(15\)
- \(16\)
- \(14\)
Show the answer and solution
Answer: E: \(14\)
Cosine rule: \(AB^2=6^2+10^2-2\cdot6\cdot10\cos120^\circ=136+60=196\), so \(AB=14\).
Question 5
In triangle \(ABC\), \(BC=7\), \(CA=15\) and angle \(ACB=60^\circ\). What is the length \(AB\)?
- \(21\)
- \(8\)
- \(13\)
- \(\sqrt{379}\)
- \(\sqrt{274}\)
Show the answer and solution
Answer: C: \(13\)
Cosine rule: \(AB^2=7^2+15^2-2\cdot7\cdot15\cos60^\circ=274-105=169\), so \(AB=13\).
Question 6
In triangle \(ABC\), \(BC=5\), \(CA=8\) and angle \(ACB=60^\circ\). What is the length \(AB\)?
- \(12\)
- \(\sqrt{129}\)
- \(3\)
- \(\sqrt{89}\)
- \(7\)
Show the answer and solution
Answer: E: \(7\)
Cosine rule: \(AB^2=5^2+8^2-2\cdot5\cdot8\cos60^\circ=89-40=49\), so \(AB=7\).
Question 7
In triangle \(ABC\), \(BC=8\), \(CA=5\) and angle \(ACB=60^\circ\). What is the length \(AB\)?
- \(\sqrt{89}\)
- \(7\)
- \(3\)
- \(12\)
- \(\sqrt{129}\)
Show the answer and solution
Answer: B: \(7\)
Cosine rule: \(AB^2=8^2+5^2-2\cdot8\cdot5\cos60^\circ=89-40=49\), so \(AB=7\).
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