Date | May 2019 | Marks available | 4 | Reference code | 19M.2.SL.TZ2.S_4 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 2 |
Command term | Find | Question number | S_4 | Adapted from | N/A |
Question
OAB is a sector of the circle with centre O and radius rr, as shown in the following diagram.
The angle AOB is θθ radians, where 0<θ<π20<θ<π2.
The point C lies on OA and OA is perpendicular to BC.
Show that OC=rcosθOC=rcosθ.
Find the area of triangle OBC in terms of rr and θ.
Given that the area of triangle OBC is 3535 of the area of sector OAB, find θ.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
cosθ=OCrcosθ=OCr A1
OC=rcosθOC=rcosθ AG N0
[1 mark]
valid approach (M1)
eg 12OC×OBsinθ , BC=rsinθ, 12rcosθ×BC , 12rsinθ×OC
area =12r2sinθcosθ (=14r2sin(2θ)) (must be in terms of r and θ) A1 N2
[2 marks]
valid attempt to express the relationship between the areas (seen anywhere) (M1)
eg OCB = 35OBA , 12r2sinθcosθ=35×12r2θ , 14r2sin2θ=310r2θ
correct equation in terms of θ only A1
eg sinθcosθ=35θ , 14sin2θ=310θ
valid attempt to solve their equation (M1)
eg sketch, −0.830017, 0
0.830017
θ = 0.830 A1 N2
Note: Do not award final A1 if additional answers given.
[4 marks]