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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 = 40°, side a (opposite A) = 5 cm, and side b (opposite B) = 7 cm. Use the sine formula to find all possible values of angle B, correct to 3 significant figures. Explain why there are two possible values.

  2. 2. In triangle ABC, angle A = 40°, side a = 10 cm, and side b = 7 cm. Find angle B, correct to 3 significant figures. Explain why only one triangle can be formed in this case.

  3. 3. In triangle ABC, angle A = 120°, side a = 15 cm, and side b = 9 cm. Solve the triangle completely: find angles B and C, and side c, all correct to 3 significant figures.

  4. 4. In triangle ABC, angle A = 120°, side a = 535\sqrt{3} cm, and side b = 9 cm. Explain why no triangle can be formed with these measurements.

  5. 5. In triangle ABC, angle B = 30° and side AB = 20 cm. Kelvin says the shortest possible length of side AC is 10 cm. Verify his claim using trigonometry and explain the geometric condition that gives this minimum.

Transcript
## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name ________________________ Class _________ 4 7. In △ABC\triangle ABC, B=30∘B = 30^\circ and AB=20AB = 20 cm. Explain (a) Kelvin claims that the minimum length of ACAC is 10 cm. Do you agree? Explain your answer. (b) Find the range of the length of ACAC such that two triangles can be formed. (a) When AC⊥BCAC \perp BC, the length of ACAC is minimum and sin⁡30∘=AC20 cm\sin 30^\circ = \frac{AC}{20 \text{ cm}} AC=20sin⁡30∘ cmAC = 20 \sin 30^\circ \text{ cm} =10 cm= 10 \text{ cm} ∴\therefore The claim is agreed. (b) When two triangles can be formed, 10 cm<AC<AB10 \text{ cm} < AC < AB 10 cm<AC<20 cm10 \text{ cm} < AC < 20 \text{ cm} **Supplementary Example 4** Level 2 (P.9.8) In each of the following, use the sine formula to find BB in △ABC\triangle ABC. (a) A=40∘A = 40^\circ, b=7b = 7 cm, a=5a = 5 cm (b) A=40∘A = 40^\circ, b=7b = 7 cm, a=10a = 10 cm (Give the answers correct to 3 significant figures.) (a) By the sine formula, 7 cmsin⁡B=5 cmsin⁡40∘\frac{7 \text{ cm}}{\sin B} = \frac{5 \text{ cm}}{\sin 40^\circ} sin⁡B=7sin⁡40∘5\sin B = \frac{7 \sin 40^\circ}{5} B=64.145∘ or 180∘−64.145∘B = 64.145^\circ \text{ or } 180^\circ - 64.145^\circ ∴\therefore B=64.1∘B = 64.1^\circ or 116∘116^\circ, cor. to 3 sig. fig. (An example of Case 3) (b) By the sine formula, 7 cmsin⁡B=10 cmsin⁡40∘\frac{7 \text{ cm}}{\sin B} = \frac{10 \text{ cm}}{\sin 40^\circ} sin⁡B=7sin⁡40∘10\sin B = \frac{7 \sin 40^\circ}{10} B=26.7∘, cor. to 3 sig. fig.B = 26.7^\circ, \text{ cor. to 3 sig. fig.} (An example of Case 4) **Supplementary Example 5** Level 2 (P.9.9) Solve △ABC\triangle ABC in each of the following. (a) A=120∘A = 120^\circ, b=9b = 9 cm, a=15a = 15 cm (b) A=120∘A = 120^\circ, b=9b = 9 cm, a=53a = 5\sqrt{3} cm (Give the answers correct to 3 significant figures.) (a) By the sine formula, 9 cmsin⁡B=15 cmsin⁡120∘\frac{9 \text{ cm}}{\sin B} = \frac{15 \text{ cm}}{\sin 120^\circ} sin⁡B=9sin⁡120∘15\sin B = \frac{9 \sin 120^\circ}{15} B=31.3∘, cor. to 3 sig. fig.B = 31.3^\circ, \text{ cor. to 3 sig. fig.} A+B+C=180∘(∠ sum of △)A + B + C = 180^\circ \quad (\angle \text{ sum of } \triangle) 120∘+31.306∘+C=180∘120^\circ + 31.306^\circ + C = 180^\circ C=28.7∘, cor. to 3 sig. fig.C = 28.7^\circ, \text{ cor. to 3 sig. fig.} By the sine formula, csin⁡28.694∘=15 cmsin⁡120∘\frac{c}{\sin 28.694^\circ} = \frac{15 \text{ cm}}{\sin 120^\circ} c=15sin⁡28.694∘sin⁡120∘ cmc = \frac{15 \sin 28.694^\circ}{\sin 120^\circ} \text{ cm} =8.32 cm, cor. to 3 sig. fig.= 8.32 \text{ cm, cor. to 3 sig. fig.} (An example of Case 6) (b) By the sine formula, 9 cmsin⁡B=53 cmsin⁡120∘\frac{9 \text{ cm}}{\sin B} = \frac{5\sqrt{3} \text{ cm}}{\sin 120^\circ} sin⁡B=9sin⁡120∘53\sin B = \frac{9 \sin 120^\circ}{5\sqrt{3}} B=64.158∘, cor. to 5 sig. fig.B = 64.158^\circ, \text{ cor. to 5 sig. fig.} A+B=120∘+64.158∘A + B = 120^\circ + 64.158^\circ =184.158∘= 184.158^\circ >180∘> 180^\circ i.e. △ABC\triangle ABC cannot be formed. ∴\therefore There is no solution. (An example of Case 5) P.4