Date | May 2021 | Marks available | 4 | Reference code | 21M.2.SL.TZ1.8 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Show that | Question number | 8 | Adapted from | N/A |
Question
Two straight fences meet at point AA and a field lies between them.
A horse is tied to a post, PP, by a rope of length rr metres. Point DD is on one fence and point EE is on the other, such that PD=PE=PA=rPD=PE=PA=r and DˆPE=θDˆPE=θ radians. This is shown in the following diagram.
The length of the arc DEDE shown in the diagram is 28 m28m.
A new fence is to be constructed between points BB and CC which will enclose the field, as shown in the following diagram.
Point CC is due west of BB and AC=800 mAC=800m . The bearing of BB from AA is 195°195°.
Write down an expression for rr in terms of θθ.
Show that the area of the field that the horse can reach is 392θ2(θ+sin θ)392θ2(θ+sinθ).
The area of field that the horse can reach is 460 m2460m2. Find the value of θθ.
Hence, find the size of DˆAEDˆAE.
Find the size of AˆBCAˆBC.
Find the length of new fence required.
Markscheme
r=28θr=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))12r2θ+12r×r×sin(π−θ) (=r22(θ+sin(π−θ)) A1
sin(π-θ)=sin θsin(π−θ)=sinθ (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]