一位农民拥有两块三角形土地 $ABC$ 和 $ADC$ ，如上图所示。已知 $AB = 7\ \text{m}$ ， $AC = 8\ \text{m}$ 且 $AD = 6\ \text{m}$ ，角 $BAD$ 为直角，角 $BAC = \theta$ 弧度。

(a)
1. (i)
  证明这两块土地的总面积 $A\ \text{m}^2$ 可以表示为 $a\sin\theta + b\cos\theta$ 的形式，其中 $a$ 和 $b$ 是整数。
2. (ii)
  将 $A$ 表示为 $R\sin(\theta+\alpha)$ 的形式，其中 $R>0$ 且 $0 < \alpha < \tfrac{\pi}{2}$ 。
3. (iii)
  求 $A$ 的最大可能值以及取得该最大值时的 $\theta$ 值。
(b)
该农民想在土地 $ABC$ 上种植玫瑰，在土地 $ADC$ 上种植木槿。已知他希望土地 $ABC$ 的面积是土地 $ADC$ 面积的两倍，求 $\theta$ 的值。

三角恒等式解三角方程 English

先不看答案，让 AI 诊断你卡在哪里？

它不会直接告诉你答案，而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

9(a)(i)

1. Express the area of triangle $ABC$ in terms of $\theta$.

Area of $\triangle ABC = \frac{1}{2}(AB)(AC)\sin(\angle BAC)$
$= \frac{1}{2}(7)(8)\sin\theta$
$= 28\sin\theta$

2. Determine the angle $CAD$ and express the area of triangle $ADC$.

Since $\angle BAD = \frac{\pi}{2}$ and $\angle BAC = \theta$ , we have $\angle CAD = \frac{\pi}{2} - \theta$ .
Area of $\triangle ADC = \frac{1}{2}(AD)(AC)\sin(\angle CAD)$
$= \frac{1}{2}(6)(8)\sin\left(\frac{\pi}{2} - \theta\right)$

3. Simplify the area of triangle $ADC$ and find the total area $A$.

Using the identity $\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$ ,
Area of $\triangle ADC = 24\cos\theta$
Total area $A = \text{Area of } \triangle ABC + \text{Area of } \triangle ADC$
$A = 28\sin\theta + 24\cos\theta$ (Shown)

答案

$A = 28\sin\theta + 24\cos\theta$

9(a)(ii)

1. Identify the expression for $A$ from part (a)(i)

From (a)(i), $A = 28\sin\theta + 24\cos\theta$ .

2. Find the value of $R$

$R = \sqrt{28^2 + 24^2} = \sqrt{784 + 576} = \sqrt{1360} \approx 36.878$

3. Find the value of $\alpha$

$\tan\alpha = \frac{24}{28} = \frac{6}{7}$
$\alpha = \tan^{-1}\left(\frac{6}{7}\right) \approx 0.7086$ radians

4. State the final expression

$A = 36.9\sin(\theta + 0.709)$ (to 3 s.f.)

答案

$A = 36.9\sin(\theta + 0.709)$

9(a)(iii)

1. State the greatest possible value of $A$ using the $R$ value from part (a)(ii)

From (a)(i) and (a)(ii), $A = 28\sin\theta + 24\cos\theta = \sqrt{1360}\sin(\theta + 0.7086)$ .
The greatest possible value of $A$ is $R$ .
$\text{Greatest } A = \sqrt{1360} \approx 36.9\ \text{m}^2$

2. Set up the equation to find the corresponding $\theta$

The maximum value occurs when $\sin(\theta + 0.7086) = 1$ .
$\theta + 0.7086 = \frac{\pi}{2}$

3. Solve for $\theta$

$\theta = \frac{\pi}{2} - 0.70862 \theta \approx 0.862\ \text{radians}$

答案

Greatest value of $A = 36.9\ \text{m}^2$ , $\theta = 0.862\ \text{radians}$

9(b)

1. Express the areas of triangles $ABC$ and $ADC$ in terms of $\theta$

Area of $ABC = \frac{1}{2}(7)(8)\sin\theta = 28\sin\theta$

Angle $DAC = \frac{\pi}{2} - \theta$
Area of $ADC = \frac{1}{2}(6)(8)\sin\left(\frac{\pi}{2} - \theta\right) = 24\cos\theta$

2. Set up the equation based on the given condition

Area of $ABC = 2 \times \text{Area of } ADC$
$28\sin\theta = 2(24\cos\theta)$
$28\sin\theta = 48\cos\theta$

3. Solve the trigonometric equation for $\theta$

$\frac{\sin\theta}{\cos\theta} = \frac{48}{28}$
$\tan\theta = \frac{12}{7}$
$\theta = \tan^{-1}\left(\frac{12}{7}\right) \approx 1.0432 \approx 1.04 \text{ radians}$

答案

$1.04$ radians

AI 分析，可能出错

用 AI 批改，圈出写错的那一步，并自动将你的薄弱点存入错题本，形成你的错因库。

一位农民拥有两块三角形土地 ABCABCABC 和 ADCADCADC，如上图所示。已知 AB=7 mAB = 7\ \text{m}AB=7 m，AC=8 mAC = 8\ \text{m}AC=8 m 且 AD=6 mAD = 6\ \text{m}AD=6 m，角 BADBADBAD 为直角，角 BAC=θBAC = \thetaBAC=θ 弧度。