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

  1. 1. In triangle ABC, AC = 11.856 cm, BC = 6 cm, and AB = 8 cm. Use the cosine formula to find ∠ACB, and explain why the answer is given to 5 significant figures.

  2. 2. In triangle ACD, AD = 6 cm, CD = 11.856 cm, and ∠ADC = 130°. Explain why ∠CAD = 22.809° is accepted but its supplement is rejected.

  3. 3. Points A, B, C, D form a quadrilateral with EC ∥ AD, E on AB. Given AB = 13 cm, DC = 7 cm, and AD = 12 cm, explain why AE = 7 cm and EC = 12 cm.

  4. 4. In triangle BCE, EB = 6 cm, BC = 9 cm, and EC = 12 cm. Use the cosine formula to find ∠ABC, and explain how EB was found.

  5. 5. In quadrilateral PQRS, ∠QPS = 59°, ∠PQS = 64°, PS = 7.5 cm, QR = 7 cm, RS = 3 cm. Explain why PQRS is not cyclic.

Transcript
## Page 1 (b) In △ABC\triangle ABC, by the cosine formula, cos⁡∠ACB=AC2+BC2−AB22×AC×BC\cos \angle ACB = \frac{AC^2 + BC^2 - AB^2}{2 \times AC \times BC} =11.8562+62−822×11.856×6= \frac{11.856^2 + 6^2 - 8^2}{2 \times 11.856 \times 6} =37.700∘, cor. to 5 sig. fig.= 37.700^\circ, \text{ cor. to } 5 \text{ sig. fig.} In △ACD\triangle ACD, by the sine formula, 6 cmsin⁡∠CAD=11.856 cmsin⁡130∘\frac{6 \text{ cm}}{\sin \angle CAD} = \frac{11.856 \text{ cm}}{\sin 130^\circ} sin⁡∠CAD=6sin⁡130∘11.856\sin \angle CAD = \frac{6 \sin 130^\circ}{11.856} ∠CAD=22.809∘ or 180−22.809∘ (rejected)\angle CAD = 22.809^\circ \text{ or } 180 - 22.809^\circ \text{ (rejected)} =22.809∘, cor. to 5 sig. fig.= 22.809^\circ, \text{ cor. to } 5 \text{ sig. fig.} ∠CAD+∠ACD+∠ADC=180∘\angle CAD + \angle ACD + \angle ADC = 180^\circ 22.809∘+∠ACD+130∘=180∘22.809^\circ + \angle ACD + 130^\circ = 180^\circ ∠ACD=27.191∘, cor. to 5 sig. fig.\angle ACD = 27.191^\circ, \text{ cor. to } 5 \text{ sig. fig.} ∠BCD=∠ACB+∠ACD\angle BCD = \angle ACB + \angle ACD =37.700∘+27.191∘= 37.700^\circ + 27.191^\circ =64.9∘, cor. to 3 sig. fig.= 64.9^\circ, \text{ cor. to } 3 \text{ sig. fig.} 25. Construct ECEC such that EC∥ADEC \parallel AD, where EE is a point on ABAB. ![Diagram: Parallelogram-like figure with points A, B, C, D, E; AB = 13 cm, AE unknown, EB unknown, AD = 12 cm, DC = 7 cm, BC = 9 cm, CD = 7 cm, angle at B indicated, EC drawn parallel to AD intersecting AB at E.] EC=AD=12 cm(opp. sides of //gram)EC = AD = 12 \text{ cm} \quad \text{(opp. sides of //gram)} AE=DC=7 cm(opp. sides of //gram)AE = DC = 7 \text{ cm} \quad \text{(opp. sides of //gram)} EB=AB−AEEB = AB - AE =(13−7) cm= (13 - 7) \text{ cm} =6 cm= 6 \text{ cm} In △BCE\triangle BCE, by the cosine formula, cos⁡∠ABC=EB2+BC2−EC22×EB×BC\cos \angle ABC = \frac{EB^2 + BC^2 - EC^2}{2 \times EB \times BC} =62+92−1222×6×9= \frac{6^2 + 9^2 - 12^2}{2 \times 6 \times 9} ∠ABC=104∘, cor. to 3 sig. fig.\angle ABC = 104^\circ, \text{ cor. to } 3 \text{ sig. fig.} 26. ∠BAD=90∘\angle BAD = 90^\circ ∠CAD=∠BAD−∠BAC\angle CAD = \angle BAD - \angle BAC =90∘−38∘= 90^\circ - 38^\circ =52∘= 52^\circ In △ACD\triangle ACD, tan⁡∠CAD=CDAD\tan \angle CAD = \frac{CD}{AD} CD=8tan⁡52∘ cmCD = 8 \tan 52^\circ \text{ cm} AB=CD=8tan⁡52∘ cmAB = CD = 8 \tan 52^\circ \text{ cm} In △ABE\triangle ABE, by the cosine formula, BE2=AB2+AE2−2×AB×AE×cos⁡∠BAEBE^2 = AB^2 + AE^2 - 2 \times AB \times AE \times \cos \angle BAE BE=(8tan⁡52∘)2+72−2×8tan⁡52∘×7×cos⁡38∘ cmBE = \sqrt{(8 \tan 52^\circ)^2 + 7^2 - 2 \times 8 \tan 52^\circ \times 7 \times \cos 38^\circ} \text{ cm} =6.39 cm, cor. to 3 sig. fig.= 6.39 \text{ cm, cor. to } 3 \text{ sig. fig.} 28. (a) In △PQS\triangle PQS, ∠PSQ+∠QPS+∠PQS=180∘\angle PSQ + \angle QPS + \angle PQS = 180^\circ ∠PSQ+59∘+64∘=180∘\angle PSQ + 59^\circ + 64^\circ = 180^\circ ∠PSQ=57∘\angle PSQ = 57^\circ By the sine formula, QSsin⁡59∘=7.5 cmsin⁡57∘\frac{QS}{\sin 59^\circ} = \frac{7.5 \text{ cm}}{\sin 57^\circ} QS=7.5sin⁡59∘sin⁡57∘ cmQS = \frac{7.5 \sin 59^\circ}{\sin 57^\circ} \text{ cm} =7.67 cm, cor. to 3 sig. fig.7.6654= 7.67 \text{ cm, cor. to } 3 \text{ sig. fig.} \quad \boxed{7.6654} (b) In △QRS\triangle QRS, by the cosine formula, cos⁡∠QRS=QR2+RS2−QS22×QR×RS\cos \angle QRS = \frac{QR^2 + RS^2 - QS^2}{2 \times QR \times RS} =72+32−7.665422×7×3= \frac{7^2 + 3^2 - 7.6654^2}{2 \times 7 \times 3} ∠QRS=91.035∘, cor. to 5 sig. fig.\angle QRS = 91.035^\circ, \text{ cor. to } 5 \text{ sig. fig.} ∠QPS+∠QRS=59∘+91.035∘\angle QPS + \angle QRS = 59^\circ + 91.035^\circ =150∘, cor. to 3 sig. fig.= 150^\circ, \text{ cor. to } 3 \text{ sig. fig.} ≠180∘\ne 180^\circ ∴PQRS is not a cyclic quadrilateral.\therefore \quad PQRS \text{ is not a cyclic quadrilateral.}