0580/41 Oct/Nov 2023, Question 5: The Cosine Rule
Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 14 marks
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The question, in short
Use right-angled triangle ACD with B on AC, BC = 54.5 m, AD = 83.2 m and angle ABD = 38° to find angle ACD; mark a circle centre on triangle EFG and explain; compare largest angles of triangles LMN and PQR.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (a) · 5 marks
Find angle ACD using BC = 54.5 m, AD = 83.2 m and angle ABD = 38° in the right-angled triangle setup.
- 1In triangle ABD: tan 38° = 83.2/AB, so AB = 83.2 ÷ tan 38° = 106.49…
- 2AC = 106.49… + 54.5 = 160.99…
- 3tan ACD = 83.2/160.99…
- 4Angle ACD = 27.32… ≈ 27.3°
Answer27.3 or 27.32 to 27.33
Part (b) · 2 marks
Mark the centre of the circle through the vertices of right-angled triangle EFG with a cross and explain your reasoning.
- 1Mark the cross at the midpoint of FG (the hypotenuse)
- 2Angle FEG = 90° and the angle in a semicircle is 90°, so FG is a diameter
AnswerCentre marked at midpoint of FG. and Angle in a semi-circle is 90
Part (c) · 7 marks
Angles L : M : N = 4 : 5 : 6; PQR has PQ = 6 cm, PR = 4 cm, QR = 5 cm. Find the difference between the largest angles.
- 1Largest angle in LMN = 6/(4 + 5 + 6) × 180 = 72°
- 2Largest angle in PQR is R, opposite PQ = 6 cm
- 3cos R = (4² + 5² − 6²)/(2 × 4 × 5) = 5/40
- 4R = 82.819…°
- 5Difference = 82.819… − 72 = 10.819… ≈ 10.8°
Answer10.8 or 10.81 to 10.82
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