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

  1. 1. In question 20, ∠ABC=72∘\angle ABC = 72^\circ and ∠BAC=60∘\angle BAC = 60^\circ. Explain why ∠DCE=∠ACB\angle DCE = \angle ACB, and find its value.

  2. 2. In question 20(b), CE=10CE = 10 cm, DE=8DE = 8 cm, and ∠DCE=48∘\angle DCE = 48^\circ. Given that ∠CDE\angle CDE is acute, use the sine formula to find ∠CDE\angle CDE correct to one decimal place.

  3. 3. In question 21, BD=CDBD = CD and ∠BCD=61∘\angle BCD = 61^\circ. Explain why ∠DBC=61∘\angle DBC = 61^\circ, and use this to find ∠ABD\angle ABD.

  4. 4. In question 21(a), ∠BAD=55∘\angle BAD = 55^\circ, ∠ABD=119∘\angle ABD = 119^\circ, and AD=16AD = 16 cm. Find ∠ADB\angle ADB and explain your reasoning using angle sum in a triangle.

  5. 5. In question 22, points PP, QQ, RR, SS lie on a circle, ∠PQR=67∘\angle PQR = 67^\circ, and ∠PSR=∠PQR\angle PSR = \angle PQR. Explain why these two angles are equal, and determine whether △PRS\triangle PRS is isosceles. Justify your answer.

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 8sin⁡a=35sin⁡103∘\frac{8}{\sin a} = \frac{3\sqrt{5}}{\sin 103^\circ} sin⁡a=1.16\sin a = 1.16 ∴No\therefore \text{No} 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. (a) ∠ACB+60∘+72∘=180∘\angle ACB + 60^\circ + 72^\circ = 180^\circ, ∠DCE=∠ACB\angle DCE = \angle ACB ∠ACB=48∘\angle ACB = 48^\circ ∠DCE=48∘\angle DCE = 48^\circ (b) 10sin⁡∠CDE=8sin⁡48∘\frac{10}{\sin \angle CDE} = \frac{8}{\sin 48^\circ} ∠CDE=68.3∘\angle CDE = 68.3^\circ 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. (a) ∠DBC=∠DCB=61∘\angle DBC = \angle DCB = 61^\circ ∠ABD=119∘\angle ABD = 119^\circ ∠ABD+55∘+∠ADB=180∘\angle ABD + 55^\circ + \angle ADB = 180^\circ ∠ADB=6∘\angle ADB = 6^\circ (b) ABsin⁡6∘=16sin⁡119∘\frac{AB}{\sin 6^\circ} = \frac{16}{\sin 119^\circ} AB=1.91AB = 1.91 cm 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. (a) ∠PSR=∠PQR=67∘\angle PSR = \angle PQR = 67^\circ 11sin⁡∠PRS=19sin⁡67∘\frac{11}{\sin \angle PRS} = \frac{19}{\sin 67^\circ} ∠PRS=32.2∘\angle PRS = 32.2^\circ ∠SPR+32.2∘+67∘=180∘\angle SPR + 32.2^\circ + 67^\circ = 180^\circ ∠SPR=80.8∘\angle SPR = 80.8^\circ (b) ∵∠SPR≠∠SRP\because \angle SPR \ne \angle SRP (by calc) ∴No\therefore \text{No}