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

  1. 1. In question 13, the sine formula is used to find angle BB, and then the angle sum of a triangle gives θ=30∘\theta = 30^\circ. Explain why θ\theta cannot be 150∘150^\circ, even though sin⁡30∘=sin⁡150∘\sin 30^\circ = \sin 150^\circ.

  2. 2. In question 14, sin⁡B≈0.8517\sin B \approx 0.8517, giving B≈45.234∘B \approx 45.234^\circ. Why is the ambiguous case (i.e., B≈180∘−45.234∘=134.766∘B \approx 180^\circ - 45.234^\circ = 134.766^\circ) rejected?

  3. 3. In question 15, solving sin⁡A=13sin⁡17∘8\sin A = \frac{13 \sin 17^\circ}{8} gives A≈28.366∘A \approx 28.366^\circ, and then θ=135∘\theta = 135^\circ. Show the full calculation for θ\theta and explain why θ\theta is obtuse here.

  4. 4. In question 16, sin⁡C≈1.04>1\sin C \approx 1.04 > 1. Explain why this means triangle ABCABC cannot be formed.

  5. 5. In question 18, sin⁡C≈2.47\sin C \approx 2.47. Compare this situation with question 16 and explain what conclusion can be drawn about triangle ABCABC.

Transcript
## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name ________________ Class ____( ) 3 13. 14. 15. θ\theta is obtuse. 13. By the sine formula, 5 cmsin⁡B=75 cmsin⁡120∘\frac{5\ \text{cm}}{\sin B} = \frac{\sqrt{75}\ \text{cm}}{\sin 120^\circ} sin⁡B=5sin⁡120∘75\sin B = \frac{5 \sin 120^\circ}{\sqrt{75}} B=30∘B = 30^\circ A+B+C=180∘A + B + C = 180^\circ 120∘+30∘+θ=180∘120^\circ + 30^\circ + \theta = 180^\circ θ=30∘\theta = 30^\circ 14. By the sine formula, 7.2 cmsin⁡B=8.6 cmsin⁡58∘\frac{7.2\ \text{cm}}{\sin B} = \frac{8.6\ \text{cm}}{\sin 58^\circ} sin⁡B=7.2sin⁡58∘8.6\sin B = \frac{7.2 \sin 58^\circ}{8.6} B=45.234∘, cor. to 5 sig. fig.B = 45.234^\circ,\ \text{cor. to 5 sig. fig.} A+B+C=180∘A + B + C = 180^\circ 58∘+45.234∘+θ=180∘58^\circ + 45.234^\circ + \theta = 180^\circ θ=76.8∘, cor. to 3 sig. fig.\theta = 76.8^\circ,\ \text{cor. to 3 sig. fig.} 15. By the sine formula, 13 cmsin⁡A=8 cmsin⁡17∘\frac{13\ \text{cm}}{\sin A} = \frac{8\ \text{cm}}{\sin 17^\circ} sin⁡A=13sin⁡17∘8\sin A = \frac{13 \sin 17^\circ}{8} A=28.366∘, cor. to 5 sig. fig.A = 28.366^\circ,\ \text{cor. to 5 sig. fig.} A+B+C=180∘A + B + C = 180^\circ 28.366∘+θ+17∘=180∘28.366^\circ + \theta + 17^\circ = 180^\circ θ=135∘, cor. to 3 sig. fig.\theta = 135^\circ,\ \text{cor. to 3 sig. fig.} In each of the following, determine whether △ABC\triangle ABC can be formed. If yes, find CC. [Nos. 16–19] 16. A=25∘A = 25^\circ, a=11a = 11 cm, c=27c = 27 cm 16. By the sine formula, 27 cmsin⁡C=11 cmsin⁡25∘\frac{27\ \text{cm}}{\sin C} = \frac{11\ \text{cm}}{\sin 25^\circ} sin⁡C=27sin⁡25∘11\sin C = \frac{27 \sin 25^\circ}{11} =1.04, cor. to 3 sig. fig.= 1.04,\ \text{cor. to 3 sig. fig.} >1> 1 ∴\therefore There is no solution for CC. ∴\therefore △ABC\triangle ABC cannot be formed. 17. A=74∘A = 74^\circ, a=16a = 16 cm, c=9c = 9 cm 17. By the sine formula, 9 cmsin⁡C=16 cmsin⁡74∘\frac{9\ \text{cm}}{\sin C} = \frac{16\ \text{cm}}{\sin 74^\circ} sin⁡C=9sin⁡74∘16\sin C = \frac{9 \sin 74^\circ}{16} C=32.7∘, cor. to 3 sig. fig.C = 32.7^\circ,\ \text{cor. to 3 sig. fig.} A+C=74∘+32.732∘A + C = 74^\circ + 32.732^\circ =107∘, cor. to 3 sig. fig.= 107^\circ,\ \text{cor. to 3 sig. fig.} <180∘< 180^\circ ∴\therefore △ABC\triangle ABC can be formed and C=32.7∘C = 32.7^\circ. 18. B=135∘B = 135^\circ, c=7c = 7 cm, b=2b = 2 cm 18. By the sine formula, 7 cmsin⁡C=2 cmsin⁡135∘\frac{7\ \text{cm}}{\sin C} = \frac{2\ \text{cm}}{\sin 135^\circ} sin⁡C=7sin⁡135∘2\sin C = \frac{7 \sin 135^\circ}{2} =2.47, cor. to 3 sig. fig.= 2.47,\ \text{cor. to 3 sig. fig.} >1> 1 ∴\therefore There is no solution for CC. ∴\therefore △ABC\triangle ABC cannot be formed. P.3