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

  1. 1. In question 19, why does the equation sin⁡A=8sin⁡103∘35≈1.16\sin A = \frac{8 \sin 103^\circ}{3\sqrt{5}} \approx 1.16 imply that triangle ABCABC cannot be formed?

  2. 2. In question 20(a), how is ∠DCE\angle DCE found to be 48∘48^\circ? Explain using angle sum in a triangle and the geometry of straight lines.

  3. 3. In question 20(b), given CE=10CE = 10 cm, DE=8DE = 8 cm, and ∠DCE=48∘\angle DCE = 48^\circ, why is ∠CDE\angle CDE taken to be acute, and how is its value calculated?

  4. 4. In question 21(a), explain why ∠ADB=6∘\angle ADB = 6^\circ, using the fact that BD=CDBD = CD and the exterior angle property.

  5. 5. In question 22(b), why is △PRS\triangle PRS not isosceles? Use the angles you know or can calculate.

Transcript
## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name _________________ Class __( ) 4 19. B=103∘B = 103^\circ, a=8a = 8 cm, b=35b = 3\sqrt{5} cm 19. By the sine formula, 8 cmsin⁡A=35 cmsin⁡103∘\frac{8 \text{ cm}}{\sin A} = \frac{3\sqrt{5} \text{ cm}}{\sin 103^\circ} sin⁡A=8sin⁡103∘35\sin A = \frac{8 \sin 103^\circ}{3\sqrt{5}} =1.16, cor. to 3 sig. fig.= 1.16, \text{ cor. to } 3 \text{ sig. fig.} >1> 1 ∴\therefore There is no solution for AA. ∴\therefore △ABC\triangle ABC cannot be formed. 20. In the figure, ACDACD and BCEBCE are two straight lines. CE=10CE = 10 cm, DE=8DE = 8 cm, ∠ABC=72∘\angle ABC = 72^\circ and ∠BAC=60∘\angle BAC = 60^\circ. (a) Find ∠DCE\angle DCE. (b) If ∠CDE\angle CDE is an acute angle, find ∠CDE\angle CDE. 20. (a) In △ABC\triangle ABC, ∠BAC+∠ABC+∠ACB=180∘\angle BAC + \angle ABC + \angle ACB = 180^\circ 60∘+72∘+∠ACB=180∘60^\circ + 72^\circ + \angle ACB = 180^\circ ∠ACB=48∘\angle ACB = 48^\circ ∠DCE=∠ACB\angle DCE = \angle ACB =48∘= 48^\circ (b) In △CDE\triangle CDE, by the sine formula, 10 cmsin⁡∠CDE=8 cmsin⁡48∘\frac{10 \text{ cm}}{\sin \angle CDE} = \frac{8 \text{ cm}}{\sin 48^\circ} sin⁡∠CDE=10sin⁡48∘8\sin \angle CDE = \frac{10 \sin 48^\circ}{8} ∠CDE=68.3∘, cor. to 3 sig. fig.\angle CDE = 68.3^\circ, \text{ cor. to } 3 \text{ sig. fig.} 21. In the figure, ABCABC is a straight line and BD=CDBD = CD. AD=16AD = 16 cm, ∠BAD=55∘\angle BAD = 55^\circ and ∠BCD=61∘\angle BCD = 61^\circ. (a) Find ∠ADB\angle ADB. (b) Find the length of ABAB. 21. (a) ∵BD=CD\because BD = CD ∴∠CBD=∠BCD=61∘\therefore \angle CBD = \angle BCD = 61^\circ In △ABD\triangle ABD, ∠ADB+∠BAD=∠CBD\angle ADB + \angle BAD = \angle CBD ∠ADB+55∘=61∘\angle ADB + 55^\circ = 61^\circ ∠ADB=6∘\angle ADB = 6^\circ (b) ∠ABD+∠CBD=180∘\angle ABD + \angle CBD = 180^\circ ∠ABD+61∘=180∘\angle ABD + 61^\circ = 180^\circ ∠ABD=119∘\angle ABD = 119^\circ In △ABD\triangle ABD, by the sine formula, ABsin⁡6∘=16 cmsin⁡119∘\frac{AB}{\sin 6^\circ} = \frac{16 \text{ cm}}{\sin 119^\circ} AB=16sin⁡6∘sin⁡119∘ cmAB = \frac{16 \sin 6^\circ}{\sin 119^\circ} \text{ cm} =1.91 cm, cor. to 3 sig. fig.= 1.91 \text{ cm}, \text{ cor. to } 3 \text{ sig. fig.} 22. In the figure, PP, QQ, RR and SS are points on a circle. PR=19PR = 19 cm, PS=11PS = 11 cm and ∠PQR=67∘\angle PQR = 67^\circ. (a) Find ∠SPR\angle SPR. Explain (b) Is △PRS\triangle PRS an isosceles triangle? Explain your answer. 22. (a) ∠PQR+∠PSR=180∘\angle PQR + \angle PSR = 180^\circ 67∘+∠PSR=180∘67^\circ + \angle PSR = 180^\circ ∠PSR=113∘\angle PSR = 113^\circ In △PRS\triangle PRS, by the sine formula, 11 cmsin⁡∠PRS=19 cmsin⁡113∘\frac{11 \text{ cm}}{\sin \angle PRS} = \frac{19 \text{ cm}}{\sin 113^\circ} sin⁡∠PRS=11sin⁡113∘19\sin \angle PRS = \frac{11 \sin 113^\circ}{19} ∠PRS=32.203∘, cor. to 5 sig. fig.\angle PRS = 32.203^\circ, \text{ cor. to } 5 \text{ sig. fig.} ∠PRS+∠SPR+∠PSR=180∘\angle PRS + \angle SPR + \angle PSR = 180^\circ 32.203∘+113∘+∠SPR=180∘32.203^\circ + 113^\circ + \angle SPR = 180^\circ ∠SPR=34.8∘, cor. to 3 sig. fig.\angle SPR = 34.8^\circ, \text{ cor. to } 3 \text{ sig. fig.} (b) The three angles in △PRS\triangle PRS are all different. ∴△PRS\therefore \triangle PRS is not an isosceles triangle. P.4