0580/33 Oct/Nov 2023, Question 9: Choosing the Right Trigonometric Ratio
Cambridge IGCSE Mathematics 0580 · Paper 33 (Core) · Calculator · 10 marks
اقرأ هذا الحل مشروحاً بالعربي ←
The question, in short
Use a semicircle with diameter AB, centre O and point C on the arc, plus two right-angled triangles PQR and XYZ, to calculate areas, a diameter and a height difference.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (ai) · 2 marks
Calculate the area of triangle ABC using AC = 9.6 cm and BC = 18 cm with the right angle at C.
- 1Area = 1/2 × 9.6 × 18 = 86.4 cm²
Answer86.4
Part (aii) · 2 marks
Show that the diameter AB of the semicircle equals 20.4 cm.
- 1Angle ACB = 90°, so AB² = 9.6² + 18²
- 2AB² = 92.16 + 324 = 416.16
- 3AB = √416.16 = 20.4 cm
Answer√(9.6² + 18²) leading to 20.4
Part (aiii) · 2 marks
Calculate the shaded area between triangle ABC and the semicircular arc.
- 1Radius = 20.4 ÷ 2 = 10.2 cm
- 2Semicircle area = 1/2 × π × 10.2² = 163.42… cm²
- 3Triangle area = 1/2 × 9.6 × 18 = 86.4 cm²
- 4Shaded area = 163.42… − 86.4 = 77.02… ≈ 77.0 cm²
Answer77.0 or 77.02 to 77.05
Part (b) · 4 marks
Calculate the difference between heights RQ and XY using PQ = 26 cm, angle QPR = 51°, XZ = 34 cm and angle XYZ = 46°.
- 1RQ = 26 × sin 51° = 20.205…
- 2tan 46° = 34/XY, so XY = 34 ÷ tan 46° = 32.833…
- 3Difference = 32.833… − 20.205… = 12.62… ≈ 12.6 cm
Answer12.6 or 12.62 to 12.63
Stuck on a step? Read the free lesson: Choosing the Right Trigonometric Ratio
Practise more choosing the right trigonometric ratio questions
Real 0580 past-paper questions, marked on your working like the exam, with a worked solution for every part.
Practise free on LemmaraMore past-paper questions on choosing the right trigonometric ratio
Also searched as: 0580/33/O/N/23 · 0580_w23_qp_33 · 0580 w23 33 · question 9.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.