O-LevelMathematicsTrigonometry4024/22 · Oct/Nov 2014 · Q912 Marks

[Figure 9.1] In triangle ABC, AB = 4m, BC = 6m and ABC = 67°. [Figure 9.2] In triangle PQR, PQ = 5m and QR = 7m. Area of triangle PQR = Area of triangle ABC. [Figure 9.3] In the parallelogram WXYZ, WX = 8m and WZ = 2m. Area of parallelogram WXYZ = Area of triangle ABC.

(a)(i)Show that the area of triangle ABC is 11.05 m² correct to 2 decimal places.[1]
(a)(ii)Find the acute angle PQR.[2]
(a)(iii)Find the obtuse angle ZWX.[3]
(b)(i)AB, AC and CD are three rods. They can be fixed together in different positions. AC = 9 cm and M is a fixed point on AB such that AM = 12cm. When CÂM = 30°, calculate CM.[3]
(b)(ii)In another position, the end D of the rod CD is fixed at the point M. CD = 12.5 cm. Calculate the increase in CÂM.[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.

[1] 11.05 confirmed
[2] 39.1° or 39.2°
[2] M1 for (1/2) × 5 × 7 × sin PQR
[3] 136.3°
[3] M1 for 8×2× sin ZWX = 1/2 × 4 × 6 × sin 67 oe and A1 for 43.7° soi or M1 for 180 - their 43.7 soi
[3] 6.16
[3] M2 for 9² + 12² – 2 × 9 × 12 × cos 30 soi or M1 for cosine formula with 1 error and A1 for 412 (soi by 20.3), 131.5 (soi by 11.5) or 117 (soi by 10.8)
[3] 41.4
[3] M2 for cos CAM = (9² + 12² - 12.5²) / (2 × 9 × 12) oe or M1 for 12.5² = 9² + 12² – 2 × 9 × 12 cos θ oe After 0, SC1 for theirA – 30, or one of M or C

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