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

  1. 1. In a triangle, two sides are a=4a = 4 cm and b=2b = 2 cm, and the angle opposite side bb is B=30∘B = 30^\circ. Use the sine formula to determine whether a triangle can be formed, and explain your conclusion.

  2. 2. Given A=40∘A = 40^\circ, c=7c = 7 cm, and a=3a = 3 cm, show that no triangle can be formed by computing sin⁡C\sin C using the sine formula and explaining why this value is impossible.

  3. 3. Triangle ABCABC has C=70∘C = 70^\circ, b=12b = 12 cm, and c=10c = 10 cm. Use the sine formula to find sin⁡B\sin B, and state whether a triangle can be formed. Justify your answer.

  4. 4. In a triangle, side a=7a = 7, side b=5b = 5, and angle A=40∘A = 40^\circ (opposite side aa). Given that θ\theta (i.e., angle BB) is acute, find θ\theta using the sine formula.

  5. 5. When two sides aa, bb and an acute angle AA (opposite side aa) are given, the number of possible triangles depends on comparing aa with h=bsin⁡Ah = b \sin A. Describe what happens in Case 3 (h<a<bh < a < b) and explain why two triangles are possible.

Transcript
## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name __________ Class __ ( ) 3 In each of the following triangles, find θ\theta. [Nos. 1–4] 1. By the sine formula, (14)sin⁡θ=(23)sin⁡(50∘)\frac{(14)}{\sin\theta} = \frac{(23)}{\sin(50^\circ)} sin⁡θ=14sin⁡50∘23\sin\theta = \frac{14\sin50^\circ}{23} θ=27.8∘\theta = 27.8^\circ 2. 3.8sin⁡θ=9sin⁡126∘\frac{3.8}{\sin\theta} = \frac{9}{\sin126^\circ} θ=20.0∘\theta = 20.0^\circ 3. θ\theta is acute. 7sin⁡θ=5sin⁡40∘\frac{7}{\sin\theta} = \frac{5}{\sin40^\circ} θ=64.1∘\theta = 64.1^\circ 4. θ\theta is obtuse. 46sin⁡θ=45sin⁡78∘\frac{46}{\sin\theta} = \frac{45}{\sin78^\circ} θ=89.1∘ (rec)90.9∘\theta = 89.1^\circ \text{ (rec)} \\ 90.9^\circ When two sides aa, bb and one acute opposite angle AA are given, the number of triangles that can be formed is 0, 1 or 2. In each figure below, let hh be the perpendicular distance from the vertex CC to its opposite side, where h=bsin⁡Ah = b\sin A. | Case 1: a<ha < h | Case 2: a=ha = h | Case 3: h<a<bh < a < b | Case 4: a≥ba \ge b | |---|---|---|---| | ![Case 1 diagram] | ![Case 2 diagram] | ![Case 3 diagram] | ![Case 4 diagram] | | No triangles can be formed. | One triangle is formed. | Two triangles are formed. | One triangle is formed. | **Instant Example 2** Determine whether △ABC\triangle ABC can be formed if A=40∘A = 40^\circ, c=7c = 7 cm and a=3a = 3 cm. By the sine formula, 7 cmsin⁡C=3 cmsin⁡40∘\frac{7\,\text{cm}}{\sin C} = \frac{3\,\text{cm}}{\sin 40^\circ} sin⁡C=7sin⁡40∘3\sin C = \frac{7\sin 40^\circ}{3} =1.50, cor. to 3 sig. fig.= 1.50,\ \text{cor. to 3 sig. fig.} >1> 1 ∴\therefore There is no solution for CC. ∵−1≤sin⁡C≤1\because -1 \le \sin C \le 1 for all values of CC. ∴△ABC\therefore \triangle ABC cannot be formed. **Instant Practice 2** Determine whether △ABC\triangle ABC can be formed if A=45∘A = 45^\circ, c=52c = 5\sqrt{2} cm and a=5a = 5 cm. By the sine formula, (52)sin⁡C=(5)sin⁡(45∘)\frac{(5\sqrt{2})}{\sin C} = \frac{(5)}{\sin(45^\circ)} sin⁡C=(52)sin⁡(45∘)(5)\sin C = \frac{(5\sqrt{2})\sin(45^\circ)}{(5)} =(1)= (1) C=(90∘)C = (90^\circ) ∴△ABC\therefore \triangle ABC (can / cannot) be formed. In each of the following, determine whether △ABC\triangle ABC can be formed. [Nos. 5–6] 5. B=30∘B = 30^\circ, a=4a = 4 cm, b=2b = 2 cm By the sine formula, (4)sin⁡A=(2)sin⁡(30∘)\frac{(4)}{\sin A} = \frac{(2)}{\sin(30^\circ)} A=90∘A = 90^\circ ∴\therefore Yes 6. C=70∘C = 70^\circ, b=12b = 12 cm, c=10c = 10 cm 10sin⁡70∘=12sin⁡B\frac{10}{\sin 70^\circ} = \frac{12}{\sin B} sin⁡θ≈1.13\sin\theta \approx 1.13 >1> 1 ∴\therefore No