← Math

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

  1. 1. In triangle ABC, angle A = 35°, angle C = 88°, and side AB = 31 cm. Explain how to find the length of side AC using the sine formula, and calculate it correct to 3 significant figures.

  2. 2. Triangle ABC has angles A = 123°, B = 22°, and side AB = 19 cm (opposite angle C). Find angle C and explain why BC is longer than AC using the sine formula and angle sizes.

  3. 3. In triangle ABC, angle A = 41°, angle C = 87°, and side AC = 22 cm (opposite angle B). Calculate the length of side BC, showing your steps using the sine formula.

  4. 4. Given 7 cmsin⁡θ=9.5 cmsin⁡61∘\frac{7 \text{ cm}}{\sin \theta} = \frac{9.5 \text{ cm}}{\sin 61^\circ}, solve for θ\theta and explain why only one acute solution is valid in this context.

  5. 5. In a triangle, side lengths 11 cm and 7 cm are opposite angles θ\theta and 37∘37^\circ respectively. Given that θ\theta is acute, find θ\theta using the sine formula and verify that the calculated value satisfies the acute condition.

Transcript
## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name ______________________ Class ____( ) Solve the following triangles. [Nos. 7–9] 7. ![](https://i.imgur.com/7.png) A+B+C=180∘A + B + C = 180^\circ 35∘+B+88∘=180∘35^\circ + B + 88^\circ = 180^\circ B=57∘‾B = \underline{57^\circ} By the sine formula, ACsin⁡57∘=31 cmsin⁡88∘\dfrac{AC}{\sin 57^\circ} = \dfrac{31 \text{ cm}}{\sin 88^\circ} AC=31sin⁡57∘sin⁡88∘ cmAC = \dfrac{31 \sin 57^\circ}{\sin 88^\circ} \text{ cm} =26.0 cm, cor. to 3 sig. fig.= 26.0 \text{ cm, cor. to 3 sig. fig.} BCsin⁡35∘=31 cmsin⁡88∘\dfrac{BC}{\sin 35^\circ} = \dfrac{31 \text{ cm}}{\sin 88^\circ} BC=31sin⁡35∘sin⁡88∘ cmBC = \dfrac{31 \sin 35^\circ}{\sin 88^\circ} \text{ cm} =17.8 cm, cor. to 3 sig. fig.= 17.8 \text{ cm, cor. to 3 sig. fig.} 8. ![](https://i.imgur.com/8.png) A+B+C=180∘A + B + C = 180^\circ 123∘+22∘+C=180∘123^\circ + 22^\circ + C = 180^\circ C=35∘‾C = \underline{35^\circ} By the sine formula, ACsin⁡22∘=19 cmsin⁡35∘\dfrac{AC}{\sin 22^\circ} = \dfrac{19 \text{ cm}}{\sin 35^\circ} AC=19sin⁡22∘sin⁡35∘ cmAC = \dfrac{19 \sin 22^\circ}{\sin 35^\circ} \text{ cm} =12.4 cm, cor. to 3 sig. fig.= 12.4 \text{ cm, cor. to 3 sig. fig.} BCsin⁡123∘=19 cmsin⁡35∘\dfrac{BC}{\sin 123^\circ} = \dfrac{19 \text{ cm}}{\sin 35^\circ} BC=19sin⁡123∘sin⁡35∘ cmBC = \dfrac{19 \sin 123^\circ}{\sin 35^\circ} \text{ cm} =27.8 cm, cor. to 3 sig. fig.= 27.8 \text{ cm, cor. to 3 sig. fig.} 9. ![](https://i.imgur.com/9.png) A+B+C=180∘A + B + C = 180^\circ 41∘+B+87∘=180∘41^\circ + B + 87^\circ = 180^\circ B=52∘‾B = \underline{52^\circ} By the sine formula, ABsin⁡87∘=22 cmsin⁡52∘\dfrac{AB}{\sin 87^\circ} = \dfrac{22 \text{ cm}}{\sin 52^\circ} AB=22sin⁡87∘sin⁡52∘ cmAB = \dfrac{22 \sin 87^\circ}{\sin 52^\circ} \text{ cm} =27.9 cm, cor. to 3 sig. fig.= 27.9 \text{ cm, cor. to 3 sig. fig.} BCsin⁡41∘=22 cmsin⁡52∘\dfrac{BC}{\sin 41^\circ} = \dfrac{22 \text{ cm}}{\sin 52^\circ} BC=22sin⁡41∘sin⁡52∘ cmBC = \dfrac{22 \sin 41^\circ}{\sin 52^\circ} \text{ cm} =18.3‾ cm, cor. to 3 sig. fig.= \underline{18.3} \text{ cm, cor. to 3 sig. fig.} In each of the following triangles, find θ\theta. [Nos. 10–15] 10. ![](https://i.imgur.com/10.png) 10. By the sine formula, 7 cmsin⁡θ=9.5 cmsin⁡61∘\dfrac{7 \text{ cm}}{\sin \theta} = \dfrac{9.5 \text{ cm}}{\sin 61^\circ} sin⁡θ=7sin⁡61∘9.5\sin \theta = \dfrac{7 \sin 61^\circ}{9.5} θ=40.1∘‾, cor. to 3 sig. fig.\theta = \underline{40.1^\circ} \text{, cor. to 3 sig. fig.} 11. ![](https://i.imgur.com/11.png) 11. By the sine formula, 15.6 cmsin⁡θ=19.3 cmsin⁡43∘\dfrac{15.6 \text{ cm}}{\sin \theta} = \dfrac{19.3 \text{ cm}}{\sin 43^\circ} sin⁡θ=15.6sin⁡43∘19.3\sin \theta = \dfrac{15.6 \sin 43^\circ}{19.3} θ=33.5∘‾, cor. to 3 sig. fig.\theta = \underline{33.5^\circ} \text{, cor. to 3 sig. fig.} 12. θ\theta is acute. ![](https://i.imgur.com/12.png) 12. By the sine formula, 11 cmsin⁡θ=7 cmsin⁡37∘\dfrac{11 \text{ cm}}{\sin \theta} = \dfrac{7 \text{ cm}}{\sin 37^\circ} sin⁡θ=11sin⁡37∘7\sin \theta = \dfrac{11 \sin 37^\circ}{7} θ=71.0∘‾, cor. to 3 sig. fig.\theta = \underline{71.0^\circ} \text{, cor. to 3 sig. fig.} P.2