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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 (where c is opposite angle C). Explain how to find angle C and then use the sine formula to write an expression for side a. Do not evaluate the expression.

  2. 2. Triangle ABC has A = 30°, C = 110°, and b = 8 cm. Explain why we can solve this triangle using the sine formula, and state which two steps (in order) must be taken.

  3. 3. In triangle ABC, A = 22°, C = 123°, and c = 30 cm. Show the calculation used to find side a, quoting the relevant part of the sine formula and substituting the given values.

  4. 4. Explain why the sine formula cannot be directly used to solve a triangle when only three angles are known (e.g., A = 50°, B = 60°, C = 70°). What essential information is missing?

  5. 5. In triangle ABC, A = 65°, C = 85°, and b = 15 cm. Calculate angle B, then write the full equation from the sine formula that lets you find side c — including numbers and units.

Transcript
## Page 1 Secondary 5 Mathematics Compulsory Part Chapter 9 Applications of Trigonometry in 2-dimensional Problems Solution 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 \blacktriangleleft \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, 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, 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. A+B+C=180∘A + B + C = 180^\circ (22∘)+B+(123∘)=180∘(22^\circ) + B + (123^\circ) = 180^\circ B=35∘B = 35^\circ By the sine formula, asin⁡A=csin⁡C\frac{a}{\sin A} = \frac{c}{\sin C} asin⁡(22∘)=(30 cm)sin⁡(123∘)\frac{a}{\sin (22^\circ)} = \frac{(30 \text{ cm})}{\sin (123^\circ)} a=(30)(sin⁡22∘)(sin⁡123∘) cma = \frac{(30)(\sin 22^\circ)}{(\sin 123^\circ)} \text{ cm} =13.4 cm, cor. to 3 sig. fig.= 13.4 \text{ cm, cor. to 3 sig. fig.} bsin⁡B=csin⁡C\frac{b}{\sin B} = \frac{c}{\sin C} bsin⁡(35∘)=(30 cm)sin⁡(123∘)\frac{b}{\sin (35^\circ)} = \frac{(30 \text{ cm})}{\sin (123^\circ)} b=(30)(sin⁡35∘)(sin⁡123∘) cmb = \frac{(30)(\sin 35^\circ)}{(\sin 123^\circ)} \text{ cm} =20.5 cm, cor. to 3 sig. fig.= 20.5 \text{ cm, cor. to 3 sig. fig.}