Revision mat · Sine & Cosine Rule

Calculator allowed14 marks

Question mix weighted by 70 real exam questions from 59 GCSE papers (2017–24), so the most common question types get the most space. Show your working.

Sine rule (find a side) $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$Sine rule (find an angle) $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}$Cosine rule (find a side) $a^2=b^2+c^2-2bc\cos A$Cosine rule (find an angle) $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$Area of a triangle $\text{Area}=\tfrac12\,ab\sin C$
Sine rule≈47% of real exam Qs
1In triangle $ABC$, angle $A = 61^\circ$, angle $B = 57^\circ$ and side $a = 10$ cm.
Work out the length of side $b$, to 1 decimal place.
Triangle ABC ()ABC10 cmb61°57°
Answer: (3 marks)
Cosine rule≈24% of real exam Qs
2In triangle $ABC$, $b = 10$ cm, $c = 9$ cm and the angle $A = 79^\circ$ between them.
Work out the length of side $a$, to 1 decimal place.
Triangle ABC ()ABCa10 cm9 cm79°
Answer: (3 marks)
Area≈11% of real exam Qs
3A triangle has two sides of length 13 cm and 9 cm with an angle of 108$^\circ$ between them.
Work out the area of the triangle, to 1 decimal place.
Triangle ABC ()ABC13 cm9 cm108°
Answer: (3 marks)
★ Exam capstonemixed & other forms ≈16% of real exam Qs
4In triangle $ABC$, angle $B = 65^\circ$, angle $C = 42^\circ$ and side $c = AB = 13$ cm.
Work out the area of triangle $ABC$, to 1 decimal place.
Triangle ABC ()ABC13 cm65°42°
Answer: (5 marks)

Mark scheme · Sine & Cosine Rule

Total: 14 marks

Award the marks shown for each correct step, then add up the total out of 14. A method mark counts even if the final answer is wrong.

1In triangle $ABC$, angle $A = 61^\circ$, angle $B = 57^\circ$ and side $a = 10$ cm.
Work out the length of side $b$, to 1 decimal place.[3 marks]
Method
Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$.
$b = \dfrac{a \sin B}{\sin A} = \dfrac{10 \times \sin 57^\circ}{\sin 61^\circ}$
$= 9.6$ cm.
Answer: $9.6$
Marks
✔1 markSine rule with correct sides/angles
✔1 markRearrange to b = a sinB / sinA
✔1 mark9.6 cm
2In triangle $ABC$, $b = 10$ cm, $c = 9$ cm and the angle $A = 79^\circ$ between them.
Work out the length of side $a$, to 1 decimal place.[3 marks]
Method
Cosine rule: $a^2 = b^2 + c^2 - 2bc\cos A$.
$a^2 = 10^2 + 9^2 - 2(10)(9)\cos 79^\circ = 146.65$.
$a = \sqrt{146.65} = 12.1$ cm.
Answer: $12.1$
Marks
✔1 markCosine rule stated with correct values
✔1 marka² = 146.7
✔1 mark12.1 cm
3A triangle has two sides of length 13 cm and 9 cm with an angle of 108$^\circ$ between them.
Work out the area of the triangle, to 1 decimal place.[3 marks]
Method
Area $= \dfrac{1}{2}ab\sin C$.
$= \dfrac{1}{2}(13)(9)\sin 108^\circ$
$= 55.6$ cm².
Answer: $55.6$
Marks
✔1 markArea = ½ab sin C
✔1 markSubstitute the two sides and included angle
✔1 mark55.6 cm²
4In triangle $ABC$, angle $B = 65^\circ$, angle $C = 42^\circ$ and side $c = AB = 13$ cm.
Work out the area of triangle $ABC$, to 1 decimal place.[5 marks]
Method
Third angle: $A = 180^\circ - 65^\circ - 42^\circ = 73^\circ$.
Sine rule for $a = BC$: $a = \dfrac{c \sin A}{\sin C} = \dfrac{13 \times \sin 73^\circ}{\sin 42^\circ} = 18.58$ cm.
Area $= \dfrac{1}{2}\,(AB)(BC)\sin B = \dfrac{1}{2}(13)(18.58)\sin 65^\circ$
$= 109.5$ cm².
Answer: $109.5$
Marks
✔1 markA = 73°
✔1 markSine rule set up for the missing side
✔1 marka = 18.6 cm
✔1 markArea = ½(AB)(BC) sin B
✔1 mark109.5 cm²
mathedup.co.uk · sheet 7LWP