一只时钟安装在建筑物屋顶的垂直墙壁上。
从时钟中心 $O$ 到分针尖端 $M$ 的距离为 15 米。
$M$ 距离屋顶的高度 $h$（米）由 $h = a \cos kt + b$ 给出，其中 $t$ 是超过整点的分钟数。$O$ 距离屋顶 20 米。屋顶视为水平面。

1. State the value of $a$ by relating it to the amplitude of the model.

The amplitude $a$ corresponds to the length of the minute hand.
$a = 15$

2. Explain the value of $b$ by relating it to the vertical shift of the model.

The parameter $b$ is the vertical shift (or mean value) of the cosine function. Physically, this is the height of the centre of the clock $O$ above the rooftop.
When the minute hand is horizontal, $\cos kt = 0$, so $h = b$. Since the centre $O$ is given to be 20 m above the rooftop, $b = 20$.

1. Identify the period of the minute hand's motion

A minute hand completes one full revolution in 60 minutes. Therefore, the period of the height function $h$ is 60 minutes.

2. Equate the period to the formula for the period of a cosine function to find $k$

The period of $\cos kt$ is given by $\frac{2\pi}{k}$.
$\frac{2\pi}{k} = 60$
$k = \frac{2\pi}{60} = \frac{\pi}{30}$

1. State the full equation of the height function using results from previous parts

From parts (a) and (b), $a = 15$, $b = 20$, and $k = \frac{\pi}{30}$.
$h = 15 \cos\left(\frac{\pi}{30} t\right) + 20$

2. Determine the coordinates of key points (maximum, minimum, and midline intersections) for the given domain

The period is $\frac{2\pi}{\pi/30} = 60$ minutes.
At $t = 0$, $h = 15(1) + 20 = 35$ (maximum).
At $t = 15$, $h = 15(0) + 20 = 20$ (midline).
At $t = 30$, $h = 15(-1) + 20 = 5$ (minimum).
At $t = 45$, $h = 15(0) + 20 = 20$ (midline).

3. Sketch the graph

Draw a smooth curve passing through $(0, 35)$, $(15, 20)$, $(30, 5)$, and $(45, 20)$ on a set of axes with $t$ on the horizontal axis and $h$ on the vertical axis.

1. Set up the equation for the height of $M$ using the parameters from earlier parts.

From parts (a) and (b), $h = 15 \cos\left(\frac{\pi}{30}t\right) + 20$.
We are given $h = 10$, so:
$15 \cos\left(\frac{\pi}{30}t\right) + 20 = 10$

2. Isolate the trigonometric function.

$15 \cos\left(\frac{\pi}{30}t\right) = -10$
$\cos\left(\frac{\pi}{30}t\right) = -\frac{2}{3}$

3. Find the basic angle and the possible values for the argument in the required interval.

Basic angle $\alpha = \cos^{-1}\left(\frac{2}{3}\right) \approx 0.84107$ rad.
Since cosine is negative, the angle lies in the 2nd and 3rd quadrants.
For the first hour, $0 \le t \le 60$, so $0 \le \frac{\pi}{30}t \le 2\pi$.
$\frac{\pi}{30}t = \pi - 0.84107 \text{ or } \pi + 0.84107$
$\frac{\pi}{30}t \approx 2.3005 \text{ or } 3.9827$

4. Solve for $t$.

$t = 2.3005 \times \frac{30}{\pi} \text{ or } t = 3.9827 \times \frac{30}{\pi}$
$t \approx 21.968 \text{ or } 38.032$
$t = 22.0 \text{ or } 38.0 \text{ (to 3 s.f.)}$