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

  1. 1. In question 16, A=25∘A = 25^\circ, a=11a = 11 cm, and c=27c = 27 cm. Explain why no triangle △ABC\triangle ABC can be formed, using the sine rule and the range of possible values for sin⁡C\sin C.

  2. 2. In question 17, A=74∘A = 74^\circ, a=16a = 16 cm, c=9c = 9 cm. Show that exactly one triangle can be formed, and find CC correct to one decimal place.

  3. 3. In question 18, B=135∘B = 135^\circ, c=7c = 7 cm, b=2b = 2 cm. Explain why no triangle can be formed, referencing the sine rule and the maximum possible value of sine.

  4. 4. In the first worked example (line 15), θ\theta is obtuse and 13sin⁡A=8sin⁡17∘\frac{13}{\sin A} = \frac{8}{\sin 17^\circ}. Given A=28.4∘A = 28.4^\circ, explain how θ=135∘\theta = 135^\circ is obtained, and verify that θ\theta is indeed obtuse.

  5. 5. Compare the outcomes of questions 16 and 18: both give sin⁡(angle)>1\sin(\text{angle}) > 1. What do their given angle–side pairs have in common that leads to this impossibility?

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. 5sin⁡B=75sin⁡120∘\frac{5}{\sin B} = \frac{\sqrt{75}}{\sin 120^\circ} B=30∘B = 30^\circ θ+30∘+120∘=180∘\theta + 30^\circ + 120^\circ = 180^\circ θ=30∘\theta = 30^\circ 7.2sin⁡θ=8.6sin⁡58∘\frac{7.2}{\sin \theta} = \frac{8.6}{\sin 58^\circ} B=45.2∘B = 45.2^\circ 5+45.2∘+58∘=180∘5 + 45.2^\circ + 58^\circ = 180^\circ θ=76.8∘\theta = 76.8^\circ 13sin⁡A=8sin⁡17∘\frac{13}{\sin A} = \frac{8}{\sin 17^\circ} A=28.4∘A = 28.4^\circ A+28.4∘+17∘=180∘A + 28.4^\circ + 17^\circ = 180^\circ θ=135∘\theta = 135^\circ 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 27sin⁡C=11sin⁡25∘\frac{27}{\sin C} = \frac{11}{\sin 25^\circ} sin⁡C=1.04\sin C = 1.04 >1> 1 ∴\therefore No 17. A=74∘A = 74^\circ, a=16a = 16 cm, c=9c = 9 cm 9sin⁡C=16sin⁡74∘\frac{9}{\sin C} = \frac{16}{\sin 74^\circ} C=32.7∘C = 32.7^\circ or 147∘147^\circ (reject) ∴\therefore Yes 18. B=135∘B = 135^\circ, c=7c = 7 cm, b=2b = 2 cm 7sin⁡C=2sin⁡135∘\frac{7}{\sin C} = \frac{2}{\sin 135^\circ} sin⁡C=2.47\sin C = 2.47 >1> 1 ∴\therefore No P.3 P.1