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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 B = 30°, angle C = 20°, and side a = 9 cm (where a is opposite angle A). Explain how to find angle A, then use the sine formula to write an expression for side b (opposite angle B). Do not evaluate the expression.

  2. 2. In triangle ABC, angle B = 72°, angle C = 48°, and side b = 13 cm. Show the steps to find side c using the sine formula, and explain why csin⁡C=bsin⁡B\frac{c}{\sin C} = \frac{b}{\sin B} is the correct ratio to use.

  3. 3. In triangle ABC, angle A = 50° and angle B = 25°. Explain why BC is not equal to 2 × AC, using the sine formula and approximate values.

  4. 4. In triangle ABC, angle A = 50°, angle B = 25°, and BC = 10 cm. Find angle C, then use the sine formula to calculate AB (to 3 significant figures). Is it possible that AB > 13 cm? Justify your answer.

  5. 5. In a triangle, two sides are 15 cm and 18 cm, and the angle opposite the 18 cm side is 35°. Use the sine formula to find the angle θ opposite the 15 cm side. Explain why θ must be acute and less than 35°.

Transcript
## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name __________ Class __ ( ) 2 1. In △ABC\triangle ABC, B=30∘B = 30^\circ, C=20∘C = 20^\circ and a=9a = 9 cm. Solve the triangle. A+B+C=180∘A+30∘+20∘=180∘A=130∘\begin{aligned} A + B + C &= 180^\circ \\ A + 30^\circ + 20^\circ &= 180^\circ \\ A &= 130^\circ \end{aligned} By the sine formula, bsin⁡B=asin⁡A\frac{b}{\sin B} = \frac{a}{\sin A} bsin⁡30∘=9 cmsin⁡130∘\frac{b}{\sin 30^\circ} = \frac{9 \text{ cm}}{\sin 130^\circ} b=9sin⁡30∘sin⁡130∘ cmb = \frac{9 \sin 30^\circ}{\sin 130^\circ} \text{ cm} =5.87 cm, cor. to 3 sig. fig.= 5.87 \text{ cm, cor. to 3 sig. fig.} csin⁡C=asin⁡A\frac{c}{\sin C} = \frac{a}{\sin A} csin⁡20∘=9 cmsin⁡130∘\frac{c}{\sin 20^\circ} = \frac{9 \text{ cm}}{\sin 130^\circ} c=9sin⁡20∘sin⁡130∘ cmc = \frac{9 \sin 20^\circ}{\sin 130^\circ} \text{ cm} =4.02 cm, cor. to 3 sig. fig.= 4.02 \text{ cm, cor. to 3 sig. fig.} 2. In △ABC\triangle ABC, B=72∘B = 72^\circ, C=48∘C = 48^\circ and b=13b = 13 cm. Solve the triangle. A+B+C=180∘A+72∘+48∘=180∘A=60∘\begin{aligned} A + B + C &= 180^\circ \\ A + 72^\circ + 48^\circ &= 180^\circ \\ A &= 60^\circ \end{aligned} By the sine formula, csin⁡C=bsin⁡B\frac{c}{\sin C} = \frac{b}{\sin B} csin⁡48∘=13 cmsin⁡72∘\frac{c}{\sin 48^\circ} = \frac{13 \text{ cm}}{\sin 72^\circ} c=13sin⁡48∘sin⁡72∘ cmc = \frac{13 \sin 48^\circ}{\sin 72^\circ} \text{ cm} =10.2 cm, cor. to 3 sig. fig.= 10.2 \text{ cm, cor. to 3 sig. fig.} asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B} asin⁡60∘=13 cmsin⁡72∘\frac{a}{\sin 60^\circ} = \frac{13 \text{ cm}}{\sin 72^\circ} a=13sin⁡60∘sin⁡72∘ cma = \frac{13 \sin 60^\circ}{\sin 72^\circ} \text{ cm} =11.8 cm, cor. to 3 sig. fig.= 11.8 \text{ cm, cor. to 3 sig. fig.} 3. In △ABC\triangle ABC, A=50∘A = 50^\circ and B=25∘B = 25^\circ. Explain (a) Is it true that BC=2ACBC = 2AC? Explain your answer. (b) If BC=10BC = 10 cm, is it possible that AB>13AB > 13 cm? Explain your answer. (a) By the sine formula, BCsin⁡A=ACsin⁡B\frac{BC}{\sin A} = \frac{AC}{\sin B} BCAC=sin⁡Asin⁡B\frac{BC}{AC} = \frac{\sin A}{\sin B} =sin⁡50∘sin⁡25∘= \frac{\sin 50^\circ}{\sin 25^\circ} =1.81, cor. to 3 sig. fig.= 1.81, \text{ cor. to 3 sig. fig.} ∵1.81≠2\because \quad 1.81 \ne 2 ∴\therefore \quad It is not true that BC=2ACBC = 2AC. (b) A+B+C=180∘50∘+25∘+C=180∘C=105∘\begin{aligned} A + B + C &= 180^\circ \\ 50^\circ + 25^\circ + C &= 180^\circ \\ C &= 105^\circ \end{aligned} By the sine formula, ABsin⁡C=BCsin⁡A\frac{AB}{\sin C} = \frac{BC}{\sin A} ABsin⁡105∘=10 cmsin⁡50∘\frac{AB}{\sin 105^\circ} = \frac{10 \text{ cm}}{\sin 50^\circ} AB=10sin⁡105∘sin⁡50∘ cmAB = \frac{10 \sin 105^\circ}{\sin 50^\circ} \text{ cm} =12.6 cm, cor. to 3 sig. fig.= 12.6 \text{ cm, cor. to 3 sig. fig.} ∵12.6 cm<13 cm\because \quad 12.6 \text{ cm} < 13 \text{ cm} ∴\therefore \quad It is not possible that AB>13AB > 13 cm. **Instant Example 1** In the figure, find θ\theta. By the sine formula, 15 cmsin⁡θ=18 cmsin⁡35∘\frac{15 \text{ cm}}{\sin \theta} = \frac{18 \text{ cm}}{\sin 35^\circ} sin⁡θ=15sin⁡35∘18\sin \theta = \frac{15 \sin 35^\circ}{18} θ=28.6∘, cor. to 3 sig. fig.\theta = 28.6^\circ, \text{ cor. to 3 sig. fig.} ∵PQ<QR\because PQ < QR ∴θ\therefore \theta must be an acute angle smaller than 35∘35^\circ. **Instant Practice 1** In the figure, find θ\theta. By the sine formula, (26 cm)sin⁡θ=(42 cm)sin⁡(64∘)\frac{(26 \text{ cm})}{\sin \theta} = \frac{(42 \text{ cm})}{\sin (64^\circ)} sin⁡θ=(26)sin⁡(64∘)(42)\sin \theta = \frac{(26) \sin (64^\circ)}{(42)} θ=33.8∘, cor. to 3 sig. fig.\theta = 33.8^\circ, \text{ cor. to 3 sig. fig.}