Date | November 2016 | Marks available | 3 | Reference code | 16N.2.AHL.TZ0.H_9 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | H_9 | Adapted from | N/A |
Question
The diagram shows two circles with centres at the points A and B and radii 2r2r and rr, respectively. The point B lies on the circle with centre A. The circles intersect at the points C and D.
Let αα be the measure of the angle CAD and θθ be the measure of the angle CBD in radians.
Find an expression for the shaded area in terms of αα, θθ and rr.
Show that α=4arcsin14α=4arcsin14.
Hence find the value of rr given that the shaded area is equal to 4.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A=2(α−sinα)r2+12(θ−sinθ)r2A=2(α−sinα)r2+12(θ−sinθ)r2 M1A1A1
Note: Award M1A1A1 for alternative correct expressions eg. A=4(α2−sinα2)r2+12θr2A=4(α2−sinα2)r2+12θr2.
[3 marks]
METHOD 1
consider for example triangle ADM where M is the midpoint of BD M1
sinα4=14sinα4=14 A1
α4=arcsin14α4=arcsin14
α=4arcsin14α=4arcsin14 AG
METHOD 2
attempting to use the cosine rule (to obtain 1−cosα2=181−cosα2=18) M1
sinα4=14sinα4=14 (obtained from sinα4=√1−cosα22sinα4=√1−cosα22) A1
α4=arcsin14α4=arcsin14
α=4arcsin14α=4arcsin14 AG
METHOD 3
sin(π2−α4)=2sinα2sin(π2−α4)=2sinα2 where θ2=π2−α4θ2=π2−α4
cosα4=4sinα4cosα4cosα4=4sinα4cosα4 M1
Note: Award M1 either for use of the double angle formula or the conversion from sine to cosine.
14=sinα414=sinα4 A1
α4=arcsin14α4=arcsin14
α=4arcsin14α=4arcsin14 AG
[2 marks]
(from triangle ADM), θ=π−α2 (=π−2arcsin14=2arcsin14=2.6362…)θ=π−α2 (=π−2arcsin14=2arcsin14=2.6362…) A1
attempting to solve 2(α−sinα)r2+12(θ−sinθ)r2=42(α−sinα)r2+12(θ−sinθ)r2=4
with α=4arcsin14α=4arcsin14 and θ=π−α2 (=2arccos14)θ=π−α2 (=2arccos14) for rr (M1)
r=1.69r=1.69 A1
[3 marks]