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

1. In question 29(a), it is given that $\triangle ABC$ is isosceles with $BC = 4\,\text{cm}$ and $BD = \frac{1}{2} \times 4\,\text{cm}$ . Explain why $BD = 2\,\text{cm}$ and how this leads to $AD = \sqrt{5}\,\text{cm}$ using Pythagoras’ theorem.
2. In question 29(a), $AE = \frac{1}{2} AD$ . Given $AD = \sqrt{5}\,\text{cm}$ , find the exact length of $ED$ . Justify your answer using the information provided.
3. Using the values $BD = 2\,\text{cm}$ and $ED = \frac{\sqrt{5}}{2}\,\text{cm}$ , apply Pythagoras’ theorem in $\triangle BDE$ to show that $BE = \frac{\sqrt{21}}{2}\,\text{cm}$ .
4. In question 29(c), $\sin\angle BAE = \frac{2}{3}$ is used. Explain how this value is obtained from $\triangle ABD$ .
5. Given $AE = \frac{\sqrt{5}}{2}\,\text{cm}$ , $BE = \frac{\sqrt{21}}{2}\,\text{cm}$ , and $\sin\angle BAE = \frac{2}{3}$ , use the sine formula in $\triangle ABE$ to verify that $\sin\theta = \frac{2\sqrt{105}}{63}$ . Show the key algebraic steps.

Transcript

## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name ______ Class ______ ( ) 2

\therefore \quad AD = EC

\therefore \quad y = 6.47

, cor. to 3 sig. fig. (opposite sides of ||gram) [6.473 4] (b) Perimeter of

ABCD

= AB + BC + CD + DA

= (18 + 9.880\,5 + 8 + 6.473\,4)\,\text{cm}

= 42.4\,\text{cm}

, cor. to 3 sig. fig. 29. (a)

\triangle ABC

is an isosceles triangle.

\therefore \quad BD = BC = \frac{1}{2} \times 4\,\text{cm} = 2\,\text{cm}

\triangle ABD

, by Pythagoras’ theorem,

AB^2 = AD^2 + BD^2

AD = \sqrt{3^2 - 2^2}\,\text{cm}

= \sqrt{5}\,\text{cm}

AE = \frac{1}{2} AD

= \frac{\sqrt{5}}{2}\,\text{cm}

ED = AE = \frac{\sqrt{5}}{2}\,\text{cm}

\triangle BDE

, by Pythagoras’ theorem,

BE^2 = BD^2 + ED^2

BE = \sqrt{2^2 + \left(\frac{\sqrt{5}}{2}\right)^2}\,\text{cm}

= \sqrt{\frac{21}{4}}\,\text{cm}

= \frac{\sqrt{21}}{2}\,\text{cm}

\triangle ABE

, by the sine formula,

\frac{AE}{\sin\theta} = \frac{BE}{\sin\angle BAE}

\sin\theta = \frac{\sqrt{5}}{2} \cdot \frac{\sin\angle BAE}{BE}

\sin\angle BAE = \frac{BD}{AB} = \frac{2}{3}

\sin\theta = \frac{\sqrt{5}}{2} \cdot \frac{\frac{2}{3}}{\frac{\sqrt{21}}{2}} = \frac{2\sqrt{5}}{3} \cdot \frac{2}{\sqrt{21}} = \frac{2\sqrt{105}}{63}