A-LevelMathematicsTrigonometry9709/12 · Oct/Nov 2018 · Q67 Marks
The diagram shows a triangle ABC in which BC = 20 cm and angle ABC = 90°. The perpendicular from B to AC meets AC at D and AD = 9 cm. Angle BCA = θ°.
- (i)By expressing the length of BD in terms of θ in each of the triangles ABD and DBC, show that 20 sin² θ = 9 cos θ.[4]
- (ii)Hence, showing all necessary working, calculate θ.[3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.
- [B1] In Δ DBC, sinθ = BD/20 → BD = 20sinθ
- [B1] In Δ ABD, tan(90-θ) = BD/9 → BD = 9cotθ
- [M1] Equates their expressions for BD and uses sinθ/cosθ = tanθ or cosθ/sinθ = cotθ if necessary.
- [A1] → 20sin²θ = 9cosθ AG
- [M1] Uses s² + c² = 1 → 20cos²θ + 9cosθ – 20 (= 0)
- [A1] → cosθ = 0.8
- [A1] → θ = 36.9° awrt
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →