下图显示了一个圆，其弦 $PQ$ 和 $RS$ 相交于点 $X$。$P, Q, R$ 和 $S$ 是圆上的点。已知 $PQ = 7\text{ cm}$，$RS = 24\text{ cm}$，角 $PXQ = 90^\circ$，且角 $PRS = \theta$，其中 $\theta$ 是变化的。

1. Find the value of $R$

$R = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$

2. Find the value of $\alpha$

$\tan \alpha = \frac{7}{24}$
$\alpha = \tan^{-1}\left(\frac{7}{24}\right) \approx 16.260^\circ$

3. Express $PR$ in the required form

$PR = 25\cos(\theta - 16.3^\circ)$

1. Find the values of $R$ and $\alpha$ from part (ii) to express $PR$ in the form $R\cos(\theta - \alpha)$

$R = \sqrt{24^2 + 7^2} = 25$
$\tan\alpha = \frac{7}{24} \implies \alpha = 16.260^\circ$
So, $PR = 25\cos(\theta - 16.260^\circ)$

2. Set up the equation for $PR = 15$ and isolate the cosine term

$25\cos(\theta - 16.260^\circ) = 15$
$\cos(\theta - 16.260^\circ) = \frac{15}{25} = 0.6$

3. Solve for $\theta$

$\theta - 16.260^\circ = \cos^{-1}(0.6)$
$\theta - 16.260^\circ = 53.130^\circ$
$\theta = 53.130^\circ + 16.260^\circ = 69.390^\circ \approx 69.4^\circ$ (to 1 d.p.)

1. State the maximum length of PR using the amplitude from the R-formula

From part (ii), $PR = 25\cos(\theta - 16.26^\circ)$.
The maximum value of $\cos(\theta - 16.26^\circ)$ is $1$.
Therefore, the maximum length of $PR$ is $25\text{ cm}$.

2. Find the corresponding value of theta

The maximum occurs when $\cos(\theta - 16.26^\circ) = 1$.
$\theta - 16.26^\circ = 0^\circ$
$\theta = 16.26^\circ \approx 16.3^\circ$ (to 1 d.p.)