Date | May 2021 | Marks available | 3 | Reference code | 21M.2.SL.TZ1.8 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Two straight fences meet at point A and a field lies between them.
A horse is tied to a post, P, by a rope of length r metres. Point D is on one fence and point E is on the other, such that PD=PE=PA=r and DˆPE=θ radians. This is shown in the following diagram.
The length of the arc DE shown in the diagram is 28 m.
A new fence is to be constructed between points B and C which will enclose the field, as shown in the following diagram.
Point C is due west of B and AC=800 m . The bearing of B from A is 195°.
Write down an expression for r in terms of θ.
Show that the area of the field that the horse can reach is 392θ2(θ+sin θ).
The area of field that the horse can reach is 460 m2. Find the value of θ.
Hence, find the size of DˆAE.
Find the size of AˆBC.
Find the length of new fence required.
Markscheme
r=28θ A1
[1 mark]
recognising sum of area of sector and area of triangle required (M1)
12r2θ+12r×r×sin(π-θ) (=r22(θ+sin(π-θ)) A1
sin(π-θ)=sin θ (substitution seen anywhere) A1
12(28θ)2θ+12(28θ)2sin θ OR 12(28θ)2(θ+sin θ) A1
area =392θ2(θ+sin θ) AG
[4 marks]
392θ2(θ+sin θ)=460 (M1)
θ=1.43917…
θ=1.44 A1
[2 marks]
π-(π-θ)2 OR θ2 (M1)
DˆAE=0.719588…
DˆAE=0.720 A1
[2 marks]
AˆBC=195-180+90 (A1)
=105° A1
[2 marks]
choosing sine rule (M1)
BCsin DˆAE=800sin105 OR BCsin DˆAE=800sin1.83 A1
BC=546 (m) A1
[2 marks]