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

  1. 1. In the figure described, ABCDABCD is a square of side 2 cm, AEBAEB is a semicircle with centre OO, and AECAEC is a quarter-circle with centre DD. Explain why △ADE\triangle ADE is isosceles.

  2. 2. Using the fact that OA=OE=1 cmOA = OE = 1\text{ cm} (since OO is the centre of semicircle AEBAEB on diameter AB=2 cmAB = 2\text{ cm}), and ∠AOE=180∘−∠ADE\angle AOE = 180^\circ - \angle ADE, show how the cosine formula applied to △AOE\triangle AOE gives AE2=2+2cos⁡∠ADEAE^2 = 2 + 2\cos \angle ADE.

  3. 3. Given DA=DE=2 cmDA = DE = 2\text{ cm} in △ADE\triangle ADE, use the cosine formula to write an expression for AE2AE^2 in terms of cos⁡∠ADE\cos \angle ADE.

  4. 4. Explain why ∠OAD+∠OED=180∘\angle OAD + \angle OED = 180^\circ implies that quadrilateral OADEOADE is cyclic.

  5. 5. Given cos⁡∠ADE=35\cos \angle ADE = \frac{3}{5}, find sin⁡∠ADE\sin \angle ADE and explain why the negative value is rejected.

Transcript
## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name: ____________ Class ( ) 4 30. In the figure, ABCDABCD is a square of side 2 cm. AEBAEB is a semi-circle with centre OO and AECAEC is a quarter of circle with centre DD. (a) Prove that OADEOADE is a cyclic quadrilateral. (b) Find cos⁡∠ADE\cos \angle ADE. 35\frac{3}{5} (c) Find the area of the shaded region. 0.962 cm20.962\,\text{cm}^2 30. (a) Join AEAE. ∠DAO=90∘\angle DAO = 90^\circ \quad (property of square) ∠DAE+∠OAE=90∘\angle DAE + \angle OAE = 90^\circ In △AOE\triangle AOE, ∵  OA=OE\because\; OA = OE ∴  ∠OEA=∠OAE\therefore\; \angle OEA = \angle OAE \quad (base ∠\angles, isos. △\triangle) In △ADE\triangle ADE, ∵  DA=DE\because\; DA = DE ∴  ∠AED=∠DAE\therefore\; \angle AED = \angle DAE \quad (base ∠\angles, isos. △\triangle) ∴  ∠OEA+∠AED=∠DAE+∠OAE\therefore\; \angle OEA + \angle AED = \angle DAE + \angle OAE ∠OED=90∘\angle OED = 90^\circ ∠OAD+∠OED\angle OAD + \angle OED \quad , cor. to =90∘+90∘= 90^\circ + 90^\circ \quad 5 sig. fig. =180∘= 180^\circ ∴  OADE\therefore\; OADE is a cyclic quadrilateral. \quad (opp. ∠\angles supp.) (b) ∠AOE+∠ADE=180∘\angle AOE + \angle ADE = 180^\circ ∠AOE=180∘−∠ADE\angle AOE = 180^\circ - \angle ADE In △AOE\triangle AOE, ∵  AB=2\because\; AB = 2 cm and O\quad O is the centre of the circle ABEABE. ∴  OA=OE=OB=1\therefore\; OA = OE = OB = 1 cm By the cosine formula, AE2=OA2+OE2−2×OA×OE×cos⁡∠AOEAE^2 = OA^2 + OE^2 - 2 \times OA \times OE \times \cos \angle AOE AE=12+12−2×1×1×cos⁡(180∘−∠ADE)AE = \sqrt{1^2 + 1^2 - 2 \times 1 \times 1 \times \cos (180^\circ - \angle ADE)} =2+2cos⁡∠ADE\quad = \sqrt{2 + 2 \cos \angle ADE} In △ADE\triangle ADE, DA=DE=2DA = DE = 2 cm By the cosine formula, AE2=DA2+DE2−2×DA×DE×cos⁡∠ADEAE^2 = DA^2 + DE^2 - 2 \times DA \times DE \times \cos \angle ADE AE=22+22−2×2×2×cos⁡∠ADEAE = \sqrt{2^2 + 2^2 - 2 \times 2 \times 2 \times \cos \angle ADE} =8−8cos⁡∠ADE\quad = \sqrt{8 - 8 \cos \angle ADE} AE=AEAE = AE 2+2cos⁡∠ADE=8−8cos⁡∠ADE\sqrt{2 + 2 \cos \angle ADE} = \sqrt{8 - 8 \cos \angle ADE} 2+2cos⁡∠ADE=8−8cos⁡∠ADE2 + 2 \cos \angle ADE = 8 - 8 \cos \angle ADE 10cos⁡∠ADE=610 \cos \angle ADE = 6 cos⁡∠ADE=35\cos \angle ADE = \frac{3}{5} (c) sin⁡2∠ADE+cos⁡2∠ADE=1\sin^2 \angle ADE + \cos^2 \angle ADE = 1 sin⁡∠ADE=±1−(35)2\sin \angle ADE = \pm \sqrt{1 - \left(\frac{3}{5}\right)^2} =45 or −45 (rejected)\quad = \frac{4}{5} \text{ or } -\frac{4}{5} \text{ (rejected)} sin⁡∠AOE\sin \angle AOE =sin⁡(180∘−∠ADE)= \sin (180^\circ - \angle ADE) =sin⁡∠ADE= \sin \angle ADE =45= \frac{4}{5} sin⁡∠AOE=45\sin \angle AOE = \frac{4}{5} ∠AOE=53.130∘ (rejected) or 126.87∘\angle AOE = 53.130^\circ \text{ (rejected) or } 126.87^\circ ∴  ∠ADE=53.130∘\therefore\; \angle ADE = 53.130^\circ, cor. to 5 sig. fig.