The diagram shows two rods PQ and QR, of lengths 12m and 5m respectively. The rods are fixed at Q such that angle PQR = $90^\circ$ and hinged at P so as to rotate in a vertical plane. The rod PQ makes an angle $\theta$ with horizontal ground.

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

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, $d = 13\cos(\theta - 22.6^\circ)$.

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 $\theta$

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