Mathematics · Oct / Nov 2020 · Paper 41 · Question 6
0580
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The question as text
North 38° 168 m 287.9 m 192 m 205.8 m D B C A NOT TO SCALE The diagram shows a field, ABCD, on horizontal ground. BC = 192 m, CD = 287.9 m, BD = 168 m and AD = 205.8 m.
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6(a)(i)[4]
106.01 to 106.02 [4]
M2 for 2 2 2 192 + 168 − 287.9 [cos[∠CBD] =] oe 2 192 168 × ×
or M1 for the implicit form
A1 for –0.276 to – 0.275
6(a)(ii)[1]
292.0 or 291.98 to 291.99 [1]
6(a)(iii)[5]
310.0 or 310.03 to 310.04 [5]
M2 for [sin A =] 168 sin(90 × − 38) 205.8
or M1 for sin A = sin(90 − 38) 168 205.8
A1 for [A =] 40.0 or 40.03 to 40.04 M1 dep for 270 + their angle DAB oe
6(b)(i)[2]
15 500 or 15 501 to 15 503. … [2]
M1 for 0.5 × 192 × 168 × sin(106) oe
6(b)(ii)[2]
55 400 [2]
FT 3.575 × their (b)(i) oe rounded to nearest 100
M1 for figs 35 75 × figs their (b)(i) or figs 554 or figs 5541 to figs 5543
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