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Mathematics · Oct / Nov 2023 · Paper 41 · Question 5

0580

As printed in the paper

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  • 5(a)[5]

    27.3 or 27.32 to 27.33 [5]

    M4 for tan[∠ACD] = 83.2 oe 83.2 tan38 + 54.5

    or M3 for [AC =] 83.2 tan38 +54.5 oe or for [CD =] 2 83.2 83.2 2 + − − 54.5 2(54.5) cos(180 38) sin38 sin38 oe

    or M2 for [AB =] 83.2 tan38 oe or for [BD =] 83.2 oe sin 38

    or M1 for tan38 = 83.2 AB oe or sin38 = 83.2 BD oe

  • 5(b)[2]

    Centre marked at midpoint of FG. and Angle in a semi-circle is 90 [3]

    B2 B1 for marking the centre at mid-point of FG

  • 5(c)[7]

    10.8 or 10.81 to 10.82 [7]

    B2 for 72 180

    or M1 for [× 6] + + 4 5 6 and, for triangle PQR

    B4 for [angle R=]82.8 or 82.81 to 82.83

    or B3 for [cosR =] 5 40 oe or better + − 4 2 5 2 6 2

    or M2 for 2 ×× 4 5

    or M1 for 62 = 42 + 52 – 2×4×5×cosR After 0 scored for triangle PQR,

    SC1 for [P =] 55.8 or 55.77 to 55.78 or [ ]41.4 = Q or 41.40 41.41 to

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