Date | November 2013 | Marks available | 6 | Reference code | 13N.2.sl.TZ0.8 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Find and Hence or otherwise | Question number | 8 | Adapted from | N/A |
Question
Consider a circle with centre O and radius 7 cm. Triangle ABC is drawn such that its vertices are on the circumference of the circle.
AB=12.2 cm, BC=10.4 cm and AˆCB=1.058 radians.
Find BˆAC.
Find AC.
Hence or otherwise, find the length of arc ABC.
Markscheme
Notes: In this question, there may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working.
Candidates may have their GDCs in degree mode, leading to incorrect answers. If working shown, award marks in line with the markscheme, with FT as appropriate.
Ignore missing or incorrect units.
evidence of choosing sine rule (M1)
eg sinˆAa=sinˆBb
correct substitution (A1)
eg sinˆA10.4=sin1.05812.2
BˆAC=0.837 A1 N2
[3 marks]
Notes: In this question, there may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working.
Candidates may have their GDCs in degree mode, leading to incorrect answers. If working shown, award marks in line with the markscheme, with FT as appropriate.
Ignore missing or incorrect units.
METHOD 1
evidence of subtracting angles from π (M1)
eg AˆBC=π−A−C
correct angle (seen anywhere) A1
AˆBC=π−1.058−0.837, 1.246, 71.4∘
attempt to substitute into cosine or sine rule (M1)
correct substitution (A1)
eg 12.22+10.42−2×12.2×10.4cos71.4, ACsin1.246=12.2sin1.058
AC=13.3 (cm) A1 N3
METHOD 2
evidence of choosing cosine rule M1
eg a2=b2+c2−2bccosA
correct substitution (A2)
eg 12.22=10.42+b2−2×10.4bcos1.058
AC=13.3 (cm) A2 N3
[5 marks]
Notes: In this question, there may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working.
Candidates may have their GDCs in degree mode, leading to incorrect answers. If working shown, award marks in line with the markscheme, with FT as appropriate.
Ignore missing or incorrect units.
METHOD 1
valid approach (M1)
eg cosAˆOC=OA2+OC2−AC22×OA×OC, AˆOC=2×AˆBC
correct working (A1)
eg 13.32=72+72−2×7×7cosAˆOC, O=2×1.246
AˆOC=2.492 (142.8∘) (A1)
EITHER
correct substitution for arc length (seen anywhere) A1
eg 2.492=l7, l=17.4, 14π×142.8360
subtracting arc from circumference (M1)
eg 2πr−l, 14π=17.4
OR
attempt to find AˆOC reflex (M1)
eg 2π−2.492, 3.79, 360−142.8
correct substitution for arc length (seen anywhere) A1
eg l=7×3.79, 14π×217.2360
THEN
arc ABC=26.5 A1 N4
METHOD 2
valid approach to find AˆOB or BˆOC (M1)
eg choosing cos rule, twice angle at circumference
correct working for finding one value, AˆOB or BˆOC (A1)
eg cosAˆOB=72+72−12.222×7×7, AˆOB=2.116,BˆOC=1.6745
two correct calculations for arc lengths
eg AB=7×2×1.058 (=14.8135), 7×1.6745 (=11.7216) (A1)(A1)
adding their arc lengths (seen anywhere)
eg rAˆOB+rBˆOC, 14.8135+11.7216, 7(2.116+1.6745) M1
arc ABC=26.5 (cm) A1 N4
Note: Candidates may work with other interior triangles using a similar method. Check calculations carefully and award marks in line with markscheme.
[6 marks]