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

  1. 1. In triangle ABC, AB = 14 cm, BC = 20 cm, and ∠C = 72°. Explain how to find the length of side AC using the cosine formula, and write the full expression (do not evaluate).

  2. 2. Triangle PQR has PQ = 15 cm, QR = 18 cm, and PR = 12 cm. Explain how to find ∠P using the cosine formula, and show the exact expression for cos⁡P\cos P before evaluating.

  3. 3. In quadrilateral ABCD, AD ∥ BC, ∠BAD = 104°, and ∠ADB = 43°. Explain why ∠DBC = 43°, and name the geometric reason.

  4. 4. In triangle ABC, AB = 8 cm, BC = 17 cm, and AC = 12 cm. Write the cosine formula expression used to find ∠ABC, and compute cos⁡∠ABC\cos \angle ABC as a simplified fraction or decimal (show working).

  5. 5. In triangle ABD, AB = 12 cm, AD = 5 cm, and ∠BAD = 90°. BD is joined. Then in triangle BCD, CD = 8 cm, BC = 9 cm, and BD is known from the first triangle. Explain how to find ∠BCD using the cosine formula, and write the full expression for cos⁡∠BCD\cos \angle BCD.

Transcript
## Page 1 # Solution of HW 9.2 19. (a) By the cosine formula, c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab \cos C c=142+202−2×14×20×cos⁡72∘ cmc = \sqrt{14^2 + 20^2 - 2 \times 14 \times 20 \times \cos 72^\circ} \text{ cm} =20.6 cm, cor. to 3 sig. fig.20.566= 20.6 \text{ cm, cor. to 3 sig. fig.} \quad \boxed{20.566} By the cosine formula, cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc} =202+20.5662−1422×20×20.566= \frac{20^2 + 20.566^2 - 14^2}{2 \times 20 \times 20.566} A=40.3∘, cor. to 3 sig. fig.40.347A = 40.3^\circ, \text{ cor. to 3 sig. fig.} \quad \boxed{40.347} A+B+C=180∘A + B + C = 180^\circ 40.347∘+B+72∘=180∘40.347^\circ + B + 72^\circ = 180^\circ B=67.7∘, cor. to 3 sig. fig.B = 67.7^\circ, \text{ cor. to 3 sig. fig.} (b) By the cosine formula, cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc} =152+182−1222×15×18= \frac{15^2 + 18^2 - 12^2}{2 \times 15 \times 18} A=41.4∘, cor. to 3 sig. fig.41.410A = 41.4^\circ, \text{ cor. to 3 sig. fig.} \quad \boxed{41.410} cos⁡B=c2+a2−b22ca\cos B = \frac{c^2 + a^2 - b^2}{2ca} =182+122−1522×18×12= \frac{18^2 + 12^2 - 15^2}{2 \times 18 \times 12} B=55.8∘, cor. to 3 sig. fig.55.771B = 55.8^\circ, \text{ cor. to 3 sig. fig.} \quad \boxed{55.771} A+B+C=180∘A + B + C = 180^\circ 41.410∘+55.771∘+C=180∘41.410^\circ + 55.771^\circ + C = 180^\circ C=82.8∘, cor. to 3 sig. fig.C = 82.8^\circ, \text{ cor. to 3 sig. fig.} 21. (a) In △ABC\triangle ABC, by the cosine formula, cos⁡∠ABC=AB2+BC2−AC22×AB×BC\cos \angle ABC = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC} =82+172−1222×8×17= \frac{8^2 + 17^2 - 12^2}{2 \times 8 \times 17} ∠ABC=39.8∘, cor. to 3 sig. fig.39.791\angle ABC = 39.8^\circ, \text{ cor. to 3 sig. fig.} \quad \boxed{39.791} (b) In △ABD\triangle ABD, by the sine formula, ADsin⁡39.791∘=8 cmsin⁡50∘\frac{AD}{\sin 39.791^\circ} = \frac{8 \text{ cm}}{\sin 50^\circ} AD=8sin⁡39.791∘sin⁡50∘ cmAD = \frac{8 \sin 39.791^\circ}{\sin 50^\circ} \text{ cm} =6.68 cm, cor. to 3 sig. fig.= 6.68 \text{ cm, cor. to 3 sig. fig.} 22. Join BDBD. In △ABD\triangle ABD, by Pythagoras’ theorem, BD2=AB2+AD2BD^2 = AB^2 + AD^2 BD=122+52 cmBD = \sqrt{12^2 + 5^2} \text{ cm} =13 cm= 13 \text{ cm} In △BCD\triangle BCD, by the cosine formula, cos⁡∠BCD=CD2+BC2−BD22×CD×BC\cos \angle BCD = \frac{CD^2 + BC^2 - BD^2}{2 \times CD \times BC} =82+92−1322×8×9= \frac{8^2 + 9^2 - 13^2}{2 \times 8 \times 9} ∠BCD=99.6∘, cor. to 3 sig. fig.\angle BCD = 99.6^\circ, \text{ cor. to 3 sig. fig.} 23. (a) In △ABD\triangle ABD, ∠ABD+∠BAD+∠ADB=180∘\angle ABD + \angle BAD + \angle ADB = 180^\circ ∠ABD+104∘+43∘=180∘\angle ABD + 104^\circ + 43^\circ = 180^\circ ∠ABD=33∘\angle ABD = 33^\circ By the sine formula, BDsin⁡104∘=6 cmsin⁡33∘\frac{BD}{\sin 104^\circ} = \frac{6 \text{ cm}}{\sin 33^\circ} BD=6sin⁡104∘sin⁡33∘ cmBD = \frac{6 \sin 104^\circ}{\sin 33^\circ} \text{ cm} =10.7 cm, cor. to 3 sig. fig.10.689= 10.7 \text{ cm, cor. to 3 sig. fig.} \quad \boxed{10.689} (b) ∠DBC=∠ADB=43∘\angle DBC = \angle ADB = 43^\circ (corr. ∠\angles, AD∥BCAD \parallel BC) In △BCD\triangle BCD, by the cosine formula, CD2=BD2+BC2−2×BD×BC×cos⁡∠DBCCD^2 = BD^2 + BC^2 - 2 \times BD \times BC \times \cos \angle DBC CD=(10.6892+152−2×10.689×15×cos⁡43∘)12 cmCD = (10.689^2 + 15^2 - 2 \times 10.689 \times 15 \times \cos 43^\circ)^{\frac{1}{2}} \text{ cm} =10.2 cm, cor. to 3 sig. fig.= 10.2 \text{ cm, cor. to 3 sig. fig.} 24. (a) BC=CD=6BC = CD = 6 cm In △ABC\triangle ABC, by the cosine formula, AC2=AB2+BC2−2×AB×BC×cos⁡∠ABCAC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos \angle ABC AC=82+62−2×8×6×cos⁡115∘ cmAC = \sqrt{8^2 + 6^2 - 2 \times 8 \times 6 \times \cos 115^\circ} \text{ cm} =11.9 cm, cor. to 3 sig. fig.11.856= 11.9 \text{ cm, cor. to 3 sig. fig.} \quad \boxed{11.856}