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S3 · 中三 questions. Answer in your own words, then submit to get a score.

  1. 1. Triangle ABC has AB = 5 cm, AC = 4 cm, and area = 8 cm². Find the acute angle A, giving your answer correct to 3 significant figures.

  2. 2. Triangle DEF is isosceles with DE = EF = 7 cm and ∠E = 100°. Calculate its area, correct to 3 significant figures.

  3. 3. The area of triangle PQR is 45 cm², PQ = 10 cm, and ∠Q = 78°. Find the length of QR, correct to 3 significant figures.

  4. 4. Triangle ABC has AC = 6 cm, BC = x cm, ∠C = 55°, and area = 20 cm². Find x, correct to 3 significant figures.

  5. 5. Triangle ABC has AC = 21 cm, BC = 18 cm, and area = 189 cm². Find ∠C, giving a reason for your answer.

Transcript
## Page 1 ① 1. In each of the following, find the area of the triangle. (a) (b) (c) 1. (a) Area of △ABC\triangle ABC =12×AB×AC×sin⁡A=12×8×15×sin⁡52∘ cm2=47.3 cm2, cor. to 3 sig. fig.= \frac{1}{2} \times AB \times AC \times \sin A \\ = \frac{1}{2} \times 8 \times 15 \times \sin 52^\circ \text{ cm}^2 \\ = 47.3 \text{ cm}^2, \text{ cor. to } 3 \text{ sig. fig.} (b) EF=DE=7EF = DE = 7 cm Area of △DEF\triangle DEF =12×DE×EF×sin⁡E=12×7×7×sin⁡100∘ cm2=24.1 cm2, cor. to 3 sig. fig.= \frac{1}{2} \times DE \times EF \times \sin E \\ = \frac{1}{2} \times 7 \times 7 \times \sin 100^\circ \text{ cm}^2 \\ = 24.1 \text{ cm}^2, \text{ cor. to } 3 \text{ sig. fig.} (c) Note that △PQR\triangle PQR is an equilateral triangle, PR=PQ=6PR = PQ = 6 cm and P=60∘P = 60^\circ. Area of △PQR\triangle PQR =12×PR×PQ×sin⁡P=12×9×9×sin⁡60∘ cm2=35.1 cm2, cor. to 3 sig. fig.= \frac{1}{2} \times PR \times PQ \times \sin P \\ = \frac{1}{2} \times 9 \times 9 \times \sin 60^\circ \text{ cm}^2 \\ = 35.1 \text{ cm}^2, \text{ cor. to } 3 \text{ sig. fig.} In each of the following, find the unknown. [Nos. 2–3] ② 2. (a) (b) (c) 2. (a) Area of △PQR=12×PQ×QR×sin⁡Q\triangle PQR = \frac{1}{2} \times PQ \times QR \times \sin Q QR=2×area of △PQRPQsin⁡Qx=2×4510sin⁡78∘ cm=9.20 cm, cor. to 3 sig. fig.QR = \frac{2 \times \text{area of } \triangle PQR}{PQ \sin Q} \\ x = \frac{2 \times 45}{10 \sin 78^\circ} \text{ cm} \\ = 9.20 \text{ cm, cor. to } 3 \text{ sig. fig.} (b) Area of △ABC=12×AC×BC×sin⁡C\triangle ABC = \frac{1}{2} \times AC \times BC \times \sin C BC=2×area of △ABCACsin⁡Cx=2×206sin⁡55∘ cm=8.14 cm, cor. to 3 sig. fig.BC = \frac{2 \times \text{area of } \triangle ABC}{AC \sin C} \\ x = \frac{2 \times 20}{6 \sin 55^\circ} \text{ cm} \\ = 8.14 \text{ cm, cor. to } 3 \text{ sig. fig.} (c) Area of △ABC=12×AC×BC×sin⁡C\triangle ABC = \frac{1}{2} \times AC \times BC \times \sin C BC=2×area of △ABCACsin⁡Cx=2×368sin⁡95∘ cm=9.03 cm, cor. to 3 sig. fig.BC = \frac{2 \times \text{area of } \triangle ABC}{AC \sin C} \\ x = \frac{2 \times 36}{8 \sin 95^\circ} \text{ cm} \\ = 9.03 \text{ cm, cor. to } 3 \text{ sig. fig.} ③ 3. (a) θ\theta is an acute angle. (b) θ\theta is an obtuse angle. (c) 3. (a) Area of △ABC=12×AB×AC×sin⁡A\triangle ABC = \frac{1}{2} \times AB \times AC \times \sin A 8=12×5×4×sin⁡θsin⁡θ=45θ=53.130∘ or 180∘−53.130∘ (reject)=53.1∘‾, cor. to 3 sig. fig.8 = \frac{1}{2} \times 5 \times 4 \times \sin \theta \\ \sin \theta = \frac{4}{5} \\ \theta = 53.130^\circ \text{ or } 180^\circ - 53.130^\circ \text{ (reject)} \\ = \underline{53.1^\circ}, \text{ cor. to } 3 \text{ sig. fig.} (b) Area of △ABC=12×AC×BC×sin⁡C\triangle ABC = \frac{1}{2} \times AC \times BC \times \sin C 50=12×12×9×sin⁡θsin⁡θ=2527θ=180∘−67.808∘ or 67.808∘ (reject)=112∘‾, cor. to 3 sig. fig.50 = \frac{1}{2} \times 12 \times 9 \times \sin \theta \\ \sin \theta = \frac{25}{27} \\ \theta = 180^\circ - 67.808^\circ \text{ or } 67.808^\circ \text{ (reject)} \\ = \underline{112^\circ}, \text{ cor. to } 3 \text{ sig. fig.} (c) Area of △ABC=12×AC×BC×sin⁡C\triangle ABC = \frac{1}{2} \times AC \times BC \times \sin C 189=12×21×18×sin⁡θsin⁡θ=1θ=90∘‾189 = \frac{1}{2} \times 21 \times 18 \times \sin \theta \\ \sin \theta = 1 \\ \theta = \underline{90^\circ}