Mathematics · Oct / Nov 2024 · Paper 42 · Question 7
0580
As printed in the paper
Show the answer and mark scheme
7(a)(i)[1]
180 – 60 – 39 [ = 81] [1]
7(a)(ii)[3]
147 or 147.1… [3]
M2 for 129sin(81) oe sin60
or M1 for sin(81) sin60 = oe CD 129
7(a)(iii)[4]
85 2 + 129 2 − 72 2 [cos = ] × × 2 85 129 31.58… [6]
M2 M1 for 72 2 = 85 2 + 129 2 −× 2 85 129cos ABD ×
A2 A1 for 0.851 to 0.852 or 9341 10965 or equivalent fraction
7(a)(iv)[3]
44.5 or 44.51 to 44.54 [3]
M2 for implicit correct method d = e.g. sin31.6 oe 85
or M1 for recognition that the line from A is perpendicular to BD
7(a)(v)[2]
247 or 247.4… [2]
M1 for 180 + (180 – 81 – 31.6) oe or for ∠NBC = 180 – 81 – 31.6 oe or for ∠NCB = 81 + 31.6 oe
7(a)(vi)[4]
972 or 973 [4]
M1 for [ABD] 1 2 × 85 × 129sin31.6 oe or 1 2 × 129 × their 44.5 oe
M1 for [BCD ] 1 2 × 129 × their147×sin39 oe their total area × 1100
M1 for 10000
7(b)[3]
126 nfww [3]
M2 for 9400 + 50 or 9400 to 9500 − 70 to 80 80 5
or M1 for 9350 or 9450 or 75 or 85 seen
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