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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°, angle B = 55°, and side c = 12 cm (side opposite angle C). Explain how to find angle C and then use the sine formula to write an expression for side a. Do not calculate the numerical value.

  2. 2. Triangle PQR has angles P = 32°, Q = 74°, and side p = 8.5 cm (opposite angle P). Describe the two main steps needed to solve this triangle completely, and state which rule is used in each step.

  3. 3. In triangle XYZ, X = 25°, Z = 110°, and y = 20 cm (side opposite angle Y). Without calculating, explain why side x must be shorter than side y.

  4. 4. In triangle ABC, A = 22°, C = 123°, and c = 30 cm. The solution shows B=35∘B = 35^\circ and uses asin⁡22∘=30sin⁡123∘\frac{a}{\sin 22^\circ} = \frac{30}{\sin 123^\circ}. Explain why asin⁡22∘=30sin⁡123∘\frac{a}{\sin 22^\circ} = \frac{30}{\sin 123^\circ} is correct, and what each part of this equation represents.

  5. 5. A student solves a triangle with A = 65°, C = 85°, and b = 15 cm, and finds B = 30°, then writes asin⁡65∘=15sin⁡30∘\frac{a}{\sin 65^\circ} = \frac{15}{\sin 30^\circ}. Explain why the denominator is sin⁡30∘\sin 30^\circ and not sin⁡65∘\sin 65^\circ or sin⁡85∘\sin 85^\circ.

Transcript
## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name 2 2 K2\ 2\ K Class 5(1)5(1) 1 Secondary 5 Mathematics Compulsory Part Chapter 9 Applications of Trigonometry in 2-dimensional Problems Teaching Examples Chapter 9.1 Sine Formula **Sine Formula** asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} Denote ∠A\angle A, ∠B\angle B and ∠C\angle C by AA, BB and CC respectively. Denote the lengths of the sides opposite to ∠A\angle A, ∠B\angle B and ∠C\angle C by aa, bb and cc respectively. **Given Two Angles and One Side (AAS or ASA)** If two angles and one side of a triangle are given, the triangle can be solved by the following steps. (i) Use the angle sum of triangle to find the remaining angle. (ii) Use the sine formula to find the other two sides. **Instant Example 1** In △ABC\triangle ABC, A=65∘A = 65^\circ, C=85∘C = 85^\circ and b=15b = 15 cm. Solve the triangle. A+B+C=180∘(∠sum of △)A + B + C = 180^\circ \quad (\angle \text{sum of } \triangle) 65∘+B+85∘=180∘65^\circ + B + 85^\circ = 180^\circ B=30∘←Step (i)B = 30^\circ \quad \leftarrow \text{Step (i)} By the sine formula, asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B} asin⁡65∘=15 cmsin⁡30∘\frac{a}{\sin 65^\circ} = \frac{15\ \text{cm}}{\sin 30^\circ} a=15sin⁡65∘sin⁡30∘ cma = \frac{15 \sin 65^\circ}{\sin 30^\circ}\ \text{cm} =27.2 cm, cor. to 3 sig. fig.= 27.2\ \text{cm},\ \text{cor. to 3 sig. fig.} csin⁡C=bsin⁡B\frac{c}{\sin C} = \frac{b}{\sin B} csin⁡85∘=15 cmsin⁡30∘\frac{c}{\sin 85^\circ} = \frac{15\ \text{cm}}{\sin 30^\circ} c=15sin⁡85∘sin⁡30∘ cmc = \frac{15 \sin 85^\circ}{\sin 30^\circ}\ \text{cm} =29.9 cm, cor. to 3 sig. fig.= 29.9\ \text{cm},\ \text{cor. to 3 sig. fig.} **Instant Practice 1** In △ABC\triangle ABC, A=22∘A = 22^\circ, C=123∘C = 123^\circ and c=30c = 30 cm. Solve the triangle. B+123∘+22∘=180∘B + 123^\circ + 22^\circ = 180^\circ (∠\angle sum) 15=35∘15 = 35^\circ asin⁡22∘=bsin⁡35∘=30sin⁡123∘\frac{a}{\sin 22^\circ} = \frac{b}{\sin 35^\circ} = \frac{30}{\sin 123^\circ} a=13.4 cma = 13.4\ \text{cm} b=20.5b = 20.5