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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 c = 7 cm (i.e., AB = 7 cm), and side a = 3 cm (i.e., BC = 3 cm). Explain why no triangle can be formed with these measurements.

  2. 2. Triangle ABC has angle A = 45°, side c = 525\sqrt{2} cm, and side a = 5 cm. Find angle C and state whether the triangle can be formed.

  3. 3. In triangle ABC, angle B = 30°, side a = 4 cm (i.e., BC), and side b = 2 cm (i.e., AC). Determine whether a triangle can be formed, and if so, find angle A.

  4. 4. In a triangle, two sides are 7 cm and 5 cm, and the angle opposite the 7 cm side is 40°. Explain why the angle opposite the 5 cm side must be acute, and find its size correct to 3 significant figures.

  5. 5. Triangle ABC has angle C = 70°, side b = 12 cm (i.e., AC), and side c = 10 cm (i.e., AB). Explain why no triangle can be formed, referring to the sine formula and the range of possible sine values.

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 cm)sin⁡θ=(23 cm)sin⁡(50∘)\frac{(14\ \text{cm})}{\sin\theta} = \frac{(23\ \text{cm})}{\sin(50^\circ)} sin⁡θ=14sin⁡50∘23\sin\theta = \frac{14\sin50^\circ}{23} θ=27.8∘‾, cor. to 3 sig. fig.\theta = \underline{27.8^\circ},\ \text{cor. to 3 sig. fig.} 2. By the sine formula, 3.8 cmsin⁡θ=9 cmsin⁡126∘\frac{3.8\ \text{cm}}{\sin\theta} = \frac{9\ \text{cm}}{\sin126^\circ} sin⁡θ=3.8sin⁡126∘9\sin\theta = \frac{3.8\sin126^\circ}{9} θ=20.0∘‾, cor. to 3 sig. fig.\theta = \underline{20.0^\circ},\ \text{cor. to 3 sig. fig.} 3. θ\theta is acute. By the sine formula, 7 cmsin⁡θ=5 cmsin⁡40∘\frac{7\ \text{cm}}{\sin\theta} = \frac{5\ \text{cm}}{\sin40^\circ} sin⁡θ=7sin⁡40∘5\sin\theta = \frac{7\sin40^\circ}{5} θ=64.1∘‾, cor. to 3 sig. fig.\theta = \underline{64.1^\circ},\ \text{cor. to 3 sig. fig.} 4. θ\theta is obtuse. By the sine formula, 46 cmsin⁡θ=45 cmsin⁡78∘\frac{46\ \text{cm}}{\sin\theta} = \frac{45\ \text{cm}}{\sin78^\circ} sin⁡θ=46sin⁡78∘45\sin\theta = \frac{46\sin78^\circ}{45} θ=180∘−89.128∘\theta = 180^\circ - 89.128^\circ =90.9∘‾, cor. to 3 sig. fig.= \underline{90.9^\circ},\ \text{cor. to 3 sig. fig.} 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](image) No triangles can be formed. | ![Case 2 diagram](image) One triangle is formed. | ![Case 3 diagram](image) Two triangles are formed. | ![Case 4 diagram](image) 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}}{\sin40^\circ} sin⁡C=7sin⁡40∘3\sin C = \frac{7\sin40^\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. ∴\therefore △ABC\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 cm)sin⁡C=(5 cm)sin⁡(45∘)\frac{(5\sqrt{2}\ \text{cm})}{\sin C} = \frac{(5\ \text{cm})}{\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) ∴\therefore △ABC\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 cm)sin⁡A=(2 cm)sin⁡(30∘)\frac{(4\ \text{cm})}{\sin A} = \frac{(2\ \text{cm})}{\sin(30^\circ)} sin⁡A=4sin⁡30∘2\sin A = \frac{4\sin30^\circ}{2} =1= 1 A=90∘A = 90^\circ ∴\therefore △ABC\triangle ABC can be formed. 6. C=70∘C = 70^\circ, b=12b = 12 cm, c=10c = 10 cm By the sine formula, 12 cmsin⁡B=10 cmsin⁡70∘\frac{12\ \text{cm}}{\sin B} = \frac{10\ \text{cm}}{\sin70^\circ} sin⁡B=12sin⁡70∘10\sin B = \frac{12\sin70^\circ}{10} =1.13, cor. to 3 sig. fig.= 1.13,\ \text{cor. to 3 sig. fig.} >1> 1 ∴\therefore There is no solution for BB. ∴\therefore △ABC\triangle ABC cannot be formed. P.3