9709 P12 - Nov 2018 - Q6
560
The diagram shows a triangle ABC in which BC = 20 cm and angle ABC is \(90^\circ\). The perpendicular from B to AC meets AC at D and AD = 9 cm. Angle BCA is \(\theta^\circ\).
- By expressing the length of BD in terms of \(\theta\) in each of the triangles ABD and DBC, show that \(20\sin^2\theta = 9\cos\theta.\)
- Hence, showing all necessary working, calculate \(\theta\).
