图中，ABAB 和 CDCD 是圆的弦，其中 AB=9 cmAB = 9\text{ cm} 且 CD=25 cmCD = 25\text{ cm}。已知 ADAD 和 BCBC 在 FF 处以 90∘90^\circ 相交，且 ∠BCD=θ\angle BCD = \theta，其中 θ\theta 是变量。

  1. (i)

    证明 AD=9cos⁡θ+25sin⁡θAD = 9 \cos \theta + 25 \sin \theta。

  2. (ii)

    将 ADAD 表示为 Rcos⁡(θ−α)R \cos(\theta - \alpha) 的形式，其中 R>0R > 0 且 0∘<α<90∘0^\circ < \alpha < 90^\circ。

  3. (iii)

    求当 AD=26 cmAD = 26\text{ cm} 时 θ\theta 的值。

  4. (iv)

    计算当 ADAD 为圆的直径时角 θ\theta 的值。

三角恒等式解三角方程平面几何证明三角函数English
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完整解法

13(i)

1. Identify angles in the same segment

Angles in the same segment are equal, so ∠BAD=∠BCD=θ\angle BAD = \angle BCD = \theta.

2. Express AD as the sum of AF and FD using trigonometry in right-angled triangles

In right-angled △ABF\triangle ABF, cos⁡∠BAF=AFAB  ⟹  AF=9cos⁡θ\cos \angle BAF = \frac{AF}{AB} \implies AF = 9 \cos \theta.
In right-angled △CDF\triangle CDF, sin⁡∠FCD=FDCD  ⟹  FD=25sin⁡θ\sin \angle FCD = \frac{FD}{CD} \implies FD = 25 \sin \theta.
Therefore, AD=AF+FD=9cos⁡θ+25sin⁡θAD = AF + FD = 9 \cos \theta + 25 \sin \theta.

答案

Shown

13(ii)

1. Expand $R \cos(\theta - \alpha)$ and compare coefficients with $9 \cos \theta + 25 \sin \theta$

Rcos⁡(θ−α)=Rcos⁡θcos⁡α+Rsin⁡θsin⁡αR \cos(\theta - \alpha) = R \cos \theta \cos \alpha + R \sin \theta \sin \alpha
Comparing with 9cos⁡θ+25sin⁡θ9 \cos \theta + 25 \sin \theta:
Rcos⁡α=9R \cos \alpha = 9
Rsin⁡α=25R \sin \alpha = 25

2. Calculate the value of $R$

R=92+252R = \sqrt{9^2 + 25^2}
R=81+625=706≈26.571R = \sqrt{81 + 625} = \sqrt{706} \approx 26.571

3. Calculate the value of $\alpha$ and write the final expression

tan⁡α=259\tan \alpha = \frac{25}{9}
α=tan⁡−1(259)≈70.201∘\alpha = \tan^{-1}\left(\frac{25}{9}\right) \approx 70.201^\circ
AD=706cos⁡(θ−70.2∘)AD = \sqrt{706} \cos(\theta - 70.2^\circ)

答案

706cos⁡(θ−70.2∘)\sqrt{706} \cos(\theta - 70.2^\circ) or 26.6cos⁡(θ−70.2∘)26.6 \cos(\theta - 70.2^\circ)

13(iii)

1. Express $AD$ using the $R$-formula result from part (ii)

From part (ii), R=92+252=706R = \sqrt{9^2 + 25^2} = \sqrt{706} and tan⁡α=259  ⟹  α≈70.201∘\tan \alpha = \frac{25}{9} \implies \alpha \approx 70.201^\circ.
So, AD=706cos⁡(θ−70.201∘)AD = \sqrt{706} \cos(\theta - 70.201^\circ).

2. Set up the equation for $AD = 26$ and find the basic angle

706cos⁡(θ−70.201∘)=26\sqrt{706} \cos(\theta - 70.201^\circ) = 26
cos⁡(θ−70.201∘)=26706\cos(\theta - 70.201^\circ) = \frac{26}{\sqrt{706}}
Basic angle =cos⁡−1(26706)≈11.903∘= \cos^{-1}\left(\frac{26}{\sqrt{706}}\right) \approx 11.903^\circ

3. Solve for the possible values of $\theta$

θ−70.201∘=11.903∘ or −11.903∘\theta - 70.201^\circ = 11.903^\circ \text{ or } -11.903^\circ
θ=70.201∘+11.903∘=82.104∘≈82.1∘\theta = 70.201^\circ + 11.903^\circ = 82.104^\circ \approx 82.1^\circ
θ=70.201∘−11.903∘=58.298∘≈58.3∘\theta = 70.201^\circ - 11.903^\circ = 58.298^\circ \approx 58.3^\circ

答案

θ=58.3∘,82.1∘\theta = 58.3^\circ, 82.1^\circ

13(iv)

1. Identify the condition for $AD$ to be the diameter

The diameter is the longest chord in a circle. Therefore, ADAD is the diameter when its length is at a maximum.

2. Find the angle $\theta$ that maximizes $AD$

From part (ii), AD=706cos⁡(θ−70.201∘)AD = \sqrt{706} \cos(\theta - 70.201^\circ).
For ADAD to be maximum, cos⁡(θ−70.201∘)=1\cos(\theta - 70.201^\circ) = 1.
θ−70.201∘=0∘\theta - 70.201^\circ = 0^\circ
θ=70.201∘≈70.2∘\theta = 70.201^\circ \approx 70.2^\circ (to 1 d.p.)

答案

70.2∘70.2^\circ

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