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

  1. 1. In Question 17, it is shown that sin⁡B=1.35\sin B = 1.35, which is greater than 1. Explain why this means triangle ABCABC cannot be formed.

  2. 2. In Question 18, after computing B=58.338∘B = 58.338^\circ, it is concluded that triangle ABCABC cannot be formed. Show the reasoning step that leads to this conclusion.

  3. 3. In Question 23(a), ∠DCB=48.590∘\angle DCB = 48.590^\circ is found using the sine formula in △ACD\triangle ACD. Given that DB=DCDB = DC, explain how ∠DBC\angle DBC is determined, and then find ∠ADB\angle ADB using angle relationships in the diagram.

  4. 4. In Question 26(a), DE∥BCDE \parallel BC is constructed in triangle ABCABC, with ∠DEA=117∘\angle DEA = 117^\circ and DE=BC=8 cmDE = BC = 8\text{ cm}. Using the sine formula in △ADE\triangle ADE, explain why only one value of ∠BAD=40.4∘\angle BAD = 40.4^\circ is accepted (and the other is rejected).

  5. 5. In Question 27, EC∥ADEC \parallel AD is drawn in quadrilateral ABCDABCD, and ∠AEC=110∘\angle AEC = 110^\circ. Explain why ∠BEC=70∘\angle BEC = 70^\circ, and use this with EB=10 cmEB = 10\text{ cm} and ∠ECB=38∘\angle ECB = 38^\circ to find ECEC using the sine formula.

Transcript
## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name _________ Class ___ ( ) 1 Solution of HW9.1 16. By the sine formula, 18 cmsin⁡B=12 cmsin⁡36∘=9 cmsin⁡42∘\frac{18 \text{ cm}}{\sin B} = \frac{12 \text{ cm}}{\sin 36^\circ} = \frac{9 \text{ cm}}{\sin 42^\circ} sin⁡B=18sin⁡36∘12=6.8\sin B = \frac{18 \sin 36^\circ}{12} = 6.8 sin⁡B=9sin⁡42∘12=6.8\sin B = \frac{9 \sin 42^\circ}{12} = 6.8 B=62.327∘ or 180∘−62.327∘B = 62.327^\circ \text{ or } 180^\circ - 62.327^\circ =62.3∘ or 118∘, cor. to 3 sig. fig.= 62.3^\circ \text{ or } 118^\circ, \text{ cor. to } 3 \text{ sig. fig.} ∴\therefore △ABC\triangle ABC can be formed with B=62.3∘B = 62.3^\circ or 118∘118^\circ. 17. By the sine formula, 5 cmsin⁡B=3 cmsin⁡54∘=5 cmsin⁡54∘\frac{5 \text{ cm}}{\sin B} = \frac{3 \text{ cm}}{\sin 54^\circ} = \frac{5 \text{ cm}}{\sin 54^\circ} sin⁡B=3sin⁡54∘5=1.35, cor. to 3 sig. fig.\sin B = \frac{3 \sin 54^\circ}{5} = 1.35, \text{ cor. to } 3 \text{ sig. fig.} >1> 1 ∴\therefore There are no solutions for BB. ∴\therefore △ABC\triangle ABC cannot be formed. 18. By the sine formula, 40 cmsin⁡B=36 cmsin⁡130∘=3640sin⁡130∘\frac{40 \text{ cm}}{\sin B} = \frac{36 \text{ cm}}{\sin 130^\circ} = \frac{36}{40 \sin 130^\circ} sin⁡B=3640sin⁡130∘\sin B = \frac{36}{40 \sin 130^\circ} B=58.338∘ or 180∘−58.338∘ (rejected)B = 58.338^\circ \text{ or } 180^\circ - 58.338^\circ \text{ (rejected)} =58.338∘, cor. to 5 sig. fig.= 58.338^\circ, \text{ cor. to } 5 \text{ sig. fig.} A+B=130∘+58.338∘=188∘, cor. to 3 sig. fig.A + B = 130^\circ + 58.338^\circ = 188^\circ, \text{ cor. to } 3 \text{ sig. fig.} >180∘> 180^\circ ∴\therefore △ABC\triangle ABC cannot be formed. 19. A+B+C=180∘A + B + C = 180^\circ A+45∘+35∘=180∘A + 45^\circ + 35^\circ = 180^\circ A=100∘A = 100^\circ By the sine formula, bsin⁡45∘=4 cmsin⁡100∘=4 cm4sin⁡45∘\frac{b}{\sin 45^\circ} = \frac{4 \text{ cm}}{\sin 100^\circ} = \frac{4 \text{ cm}}{4 \sin 45^\circ} b=4sin⁡45∘sin⁡100∘ cmb = \frac{4 \sin 45^\circ}{\sin 100^\circ} \text{ cm} =2.87 cm, cor. to 3 sig. fig.= 2.87 \text{ cm, cor. to } 3 \text{ sig. fig.} By the sine formula, csin⁡35∘=4 cmsin⁡100∘\frac{c}{\sin 35^\circ} = \frac{4 \text{ cm}}{\sin 100^\circ} c=4sin⁡35∘sin⁡100∘ cmc = \frac{4 \sin 35^\circ}{\sin 100^\circ} \text{ cm} =2.33 cm, cor. to 3 sig. fig.= 2.33 \text{ cm, cor. to } 3 \text{ sig. fig.} 20. By the sine formula, ABsin⁡18.590∘=6 cmsin⁡30∘\frac{AB}{\sin 18.590^\circ} = \frac{6 \text{ cm}}{\sin 30^\circ} AB=6sin⁡18.590∘ cmAB = 6 \sin 18.590^\circ \text{ cm} sin⁡30∘\sin 30^\circ =3.83 cm, cor. to 3 sig. fig.= 3.83 \text{ cm, cor. to } 3 \text{ sig. fig.} 23. (a) In △ACD\triangle ACD, by the sine formula, 9 cmsin⁡∠DCB=6 cmsin⁡30∘\frac{9 \text{ cm}}{\sin \angle DCB} = \frac{6 \text{ cm}}{\sin 30^\circ} sin⁡∠DCB=9sin⁡30∘6\sin \angle DCB = \frac{9 \sin 30^\circ}{6} ∠DCB=48.590∘, cor. to 5 sig. fig.\angle DCB = 48.590^\circ, \text{ cor. to } 5 \text{ sig. fig.} ∴\therefore DB=DCDB = DC ∴\therefore ∠DBC=∠DCB=48.590∘, cor. to 5 sig. fig.\angle DBC = \angle DCB = 48.590^\circ, \text{ cor. to } 5 \text{ sig. fig.} In △ADB\triangle ADB, ∠ADB+∠BAD=∠DBC\angle ADB + \angle BAD = \angle DBC ∠ADB+30∘=48.590∘\angle ADB + 30^\circ = 48.590^\circ ∠ADB=18.6∘, cor. to 3 sig. fig.[18.590]\angle ADB = 18.6^\circ, \text{ cor. to } 3 \text{ sig. fig.} \quad [18.590] (b) In △ABD\triangle ABD, by the sine formula, ABsin⁡18.590∘=6 cmsin⁡30∘\frac{AB}{\sin 18.590^\circ} = \frac{6 \text{ cm}}{\sin 30^\circ} AB=6sin⁡18.590∘ cmAB = 6 \sin 18.590^\circ \text{ cm} sin⁡30∘\sin 30^\circ =3.83 cm, cor. to 3 sig. fig.= 3.83 \text{ cm, cor. to } 3 \text{ sig. fig.} 25. (a) Construct EAEA such that EA∥CBEA \parallel CB, where EE is a point on DCDC. [Diagram: quadrilateral ABCD with AB = 3.5 cm, BC = 3.6 cm, angle at C = 72°, angle at D = 66°, and EA parallel to CB, intersecting DC at E.] ∠AED=∠BCD=72∘(corr. ∠s,EA∥CB)\angle AED = \angle BCD = 72^\circ \quad (\text{corr. } \angle s, EA \parallel CB) AE=BC=3.6 cm(opp. sides of //gram)AE = BC = 3.6 \text{ cm} \quad (\text{opp. sides of //gram}) In △ADE\triangle ADE, by the sine formula, ADsin⁡72∘=AEsin⁡65∘\frac{AD}{\sin 72^\circ} = \frac{AE}{\sin 65^\circ} AD=3.6sin⁡72∘sin⁡65∘ cmAD = \frac{3.6 \sin 72^\circ}{\sin 65^\circ} \text{ cm} =3.78 cm, cor. to 3 sig. fig.= 3.78 \text{ cm, cor. to } 3 \text{ sig. fig.} (b) In △ADE\triangle ADE, ∠DAE+∠AED+∠ADC=180∘\angle DAE + \angle AED + \angle ADC = 180^\circ ∠DAE+72∘+65∘=180∘\angle DAE + 72^\circ + 65^\circ = 180^\circ ∠DAE=43∘\angle DAE = 43^\circ By the sine formula, AEsin⁡43∘=DEsin⁡65∘\frac{AE}{\sin 43^\circ} = \frac{DE}{\sin 65^\circ} DE=3.6sin⁡43∘sin⁡65∘ cmDE = \frac{3.6 \sin 43^\circ}{\sin 65^\circ} \text{ cm} =2.709 cm, cor. to 5 sig. fig.= 2.709 \text{ cm, cor. to } 5 \text{ sig. fig.} EC=AB−AE=(18−8) cm=10 cmEC = AB - AE = (18 - 8) \text{ cm} = 10 \text{ cm} DC=DE+EC=(2.709+3.5) cm=6.21 cm, cor. to 3 sig. fig.DC = DE + EC = (2.709 + 3.5) \text{ cm} = 6.21 \text{ cm, cor. to } 3 \text{ sig. fig.} 26. (a) Construct DEDE such that DE∥BCDE \parallel BC, where EE is a point on ABAB. [Diagram: triangle ABC with AB = 11 cm, BC = 8 cm, angle at B = 117°, and DE parallel to BC, intersecting AB at E and AC at D.] ∠DEA=∠CBA=117∘(corr. ∠s,DE∥BC)\angle DEA = \angle CBA = 117^\circ \quad (\text{corr. } \angle s, DE \parallel BC) DE=BC=8 cm(opp. sides of //gram)DE = BC = 8 \text{ cm} \quad (\text{opp. sides of //gram}) In △ADE\triangle ADE, by the sine formula, 8 cmsin⁡∠BAD=11 cmsin⁡117∘\frac{8 \text{ cm}}{\sin \angle BAD} = \frac{11 \text{ cm}}{\sin 117^\circ} sin⁡∠BAD=8sin⁡117∘11\sin \angle BAD = \frac{8 \sin 117^\circ}{11} ∠BAD=40.391∘ or 180∘−40.391∘ (rejected)\angle BAD = 40.391^\circ \text{ or } 180^\circ - 40.391^\circ \text{ (rejected)} =40.4∘, cor. to 3 sig. fig.[40.391]= 40.4^\circ, \text{ cor. to } 3 \text{ sig. fig.} \quad [40.391] 27. (a) Construct ECEC such that EC∥ADEC \parallel AD, where EE is a point on ABAB. [Diagram: quadrilateral ABCD with AB = 8 cm, BC = 8 cm, CD = 8 cm, DA = 8 cm? Actually labeled: AB = 8 cm, BC = ?, CD = ?, DA = ?, angles: at D = 110°, at C = 38°, at B = ?, at A = ?, with EC parallel to AD, E on AB, and AE = ?] Given: AE=DC=8 cmAE = DC = 8 \text{ cm} EB=AB−AE=(18−8) cm=10 cmEB = AB - AE = (18 - 8) \text{ cm} = 10 \text{ cm} ∠AEC=∠ADC=110∘(opp. ∠s of //gram)\angle AEC = \angle ADC = 110^\circ \quad (\text{opp. } \angle s \text{ of //gram}) ∠BEC+∠AEC=180∘\angle BEC + \angle AEC = 180^\circ ∠BEC+110∘=180∘\angle BEC + 110^\circ = 180^\circ ∠BEC=70∘\angle BEC = 70^\circ In △BCE\triangle BCE, ∠BCE+∠BEC+∠CBE=180∘\angle BCE + \angle BEC + \angle CBE = 180^\circ ∠BCE+70∘+38∘=180∘\angle BCE + 70^\circ + 38^\circ = 180^\circ ∠BCE=72∘\angle BCE = 72^\circ By the sine formula, x cmsin⁡70∘=10 cmsin⁡72∘\frac{x \text{ cm}}{\sin 70^\circ} = \frac{10 \text{ cm}}{\sin 72^\circ} x=10sin⁡70∘sin⁡72∘x = \frac{10 \sin 70^\circ}{\sin 72^\circ} =9.88, cor. to 3 sig. fig.[9.8805]= 9.88, \text{ cor. to } 3 \text{ sig. fig.} \quad [9.8805] EC=10 cmsin⁡38∘⋅sin⁡72∘EC = \frac{10 \text{ cm}}{\sin 38^\circ} \cdot \sin 72^\circ EC=10sin⁡38∘sin⁡72∘ cmEC = \frac{10 \sin 38^\circ}{\sin 72^\circ} \text{ cm} =6.473 cm, cor. to 5 sig. fig.= 6.473 \text{ cm, cor. to } 5 \text{ sig. fig.} S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name _________ Class ___ ( ) 1 25. (a) Construct EAEA such that EA∥CBEA \parallel CB, where EE is a point on DCDC. [Same diagram as above] ∠DAE+∠AED+∠AED=180∘\angle DAE + \angle AED + \angle AED = 180^\circ Wait — correction from earlier part: Actually in original page, it says: ∠DAE+∠AED+∠AED=180∘(typo? should be ∠DAE+∠AED+∠ADE=180∘)\angle DAE + \angle AED + \angle AED = 180^\circ \quad \text{(typo? should be } \angle DAE + \angle AED + \angle ADE = 180^\circ\text{)} But as written on page: ∠DAE+∠AED+∠AED=180∘\angle DAE + \angle AED + \angle AED = 180^\circ No — looking again: the printed text says: **(b)** In △ADE\triangle ADE, ∠DAE+∠AED+∠ADE=180∘\angle DAE + \angle AED + \angle ADE = 180^\circ 40.391∘+∠ADE+117∘=180∘40.391^\circ + \angle ADE + 117^\circ = 180^\circ ∠ADE=22.609∘, cor. to 5 sig. fig.\angle ADE = 22.609^\circ, \text{ cor. to } 5 \text{ sig. fig.} By the sine formula, AEsin⁡∠ADE=11 cmsin⁡117∘\frac{AE}{\sin \angle ADE} = \frac{11 \text{ cm}}{\sin 117^\circ} AE=11sin⁡22.609∘sin⁡117∘ cmAE = \frac{11 \sin 22.609^\circ}{\sin 117^\circ} \text{ cm} =4.7461 cm, cor. to 5 sig. fig.= 4.7461 \text{ cm, cor. to } 5 \text{ sig. fig.} BE=CD=8 cm(opp. sides of //gram)BE = CD = 8 \text{ cm} \quad (\text{opp. sides of //gram}) AB=AE+BE=(4.7461+8) cm=12.7 cm, cor. to 3 sig. fig.AB = AE + BE = (4.7461 + 8) \text{ cm} = 12.7 \text{ cm, cor. to } 3 \text{ sig. fig.} [Note: The page contains repeated problems (e.g., 25 appears twice with slightly different labeling), but per instruction, transcribe exactly as written.] At bottom right: P.1