图中显示了两根杆 PQ 和 QR，长度分别为 12m 和 5m。两根杆固定在 Q 点，使得角 PQR = $90^\circ$ ，并在 P 点通过铰链连接，以便在垂直平面内旋转。杆 PQ 与水平地面成角 $\theta$ 。

(a)
用 $\theta$ 表示 $d$ 的表达式，其中 $d$ 是 R 离 P 的水平距离。
(b)
将 $d$ 表示为 $R\cos(\theta-\alpha)$ 的形式，其中 $R>0$ 且 $0^\circ<\alpha<90^\circ$ 。
(c)
已知 $d=10$ ，求 $\theta$ 的值。

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完整解法

6(a)

1. Identify the horizontal components of PQ and QR

Let the horizontal distance from P to the vertical line through Q be $d_1$ , and the horizontal distance from the vertical line through Q to the vertical line through R be $d_2$ .
Then $d = d_1 + d_2$ .

2. Express the horizontal component of PQ in terms of $\theta$

In the right-angled triangle formed by PQ and the horizontal, the horizontal distance is $d_1 = PQ \cos\theta = 12\cos\theta$ .

3. Express the horizontal component of QR in terms of $\theta$ and find $d$

The angle that QR makes with the vertical is $\theta$ (since $\angle PQR = 90^\circ$ ).
Thus, the horizontal distance $d_2 = QR \sin\theta = 5\sin\theta$ .
Therefore, $d = 12\cos\theta + 5\sin\theta$ .

答案

$d = 12\cos\theta + 5\sin\theta$

6(b)

1. State the expression for $d$ from part (a) and expand $R\cos(\theta-\alpha)$

From part (a), $d = 12\cos\theta + 5\sin\theta$ .
We want to express this in the form $R\cos(\theta-\alpha)$ .
$R\cos(\theta-\alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$

2. Compare coefficients to find equations for $R$ and $\alpha$

Comparing coefficients with $12\cos\theta + 5\sin\theta$ :
$R\cos\alpha = 12$
$R\sin\alpha = 5$

3. Calculate the value of $R$

$R = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$

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

$\tan\alpha = \frac{5}{12}$
$\alpha = \tan^{-1}\left(\frac{5}{12}\right) \approx 22.619^\circ \approx 22.6^\circ$ (to 1 d.p.)
Therefore, $d = 13\cos(\theta - 22.6^\circ)$ .

答案

$13\cos(\theta - 22.6^\circ)$

6(c)

1. Recall the expression for $d$ from parts (a) and (b) and set up the equation

From (a), $d = 12\cos\theta + 5\sin\theta$ .
From (b), $d = 13\cos(\theta - 22.62^\circ)$ .
Given $d = 10$ , we have:
$13\cos(\theta - 22.62^\circ) = 10$

2. Solve for $\theta$

$\cos(\theta - 22.62^\circ) = \frac{10}{13}$
$\theta - 22.62^\circ = \cos^{-1}\left(\frac{10}{13}\right)$
$\theta - 22.62^\circ = 39.715^\circ$
$\theta = 39.715^\circ + 22.62^\circ = 62.335^\circ \approx 62.3^\circ$

答案

$17.2^\circ$

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图中显示了两根杆 PQ 和 QR，长度分别为 12m 和 5m。两根杆固定在 Q 点，使得角 PQR = 90∘90^\circ90∘，并在 P 点通过铰链连接，以便在垂直平面内旋转。杆 PQ 与水平地面成角 θ\thetaθ。

完整解法

6(a)

6(b)

6(c)

同一知识点的其他题

图中显示了两根杆 PQ 和 QR，长度分别为 12m 和 5m。两根杆固定在 Q 点，使得角 PQR = $90^\circ$ ，并在 P 点通过铰链连接，以便在垂直平面内旋转。杆 PQ 与水平地面成角 $\theta$ 。