(a) \(\theta = 2.08\) radians (M1A1A1, N3).

Working:
Recognize that 45 = 10 + 10 + arc length (M1).
Arc length = 25 cm (A1).
Arc length formula: \( r \theta = 25 \), with \( r = 12 \), so \( 12 \theta = 25 \).
Solve: \( \theta = \frac{25}{12} \approx 2.0833 \).
Rounded to three significant figures: \( \theta \approx 2.08 \) (A1).

[3 marks]

(b) Area of the cross-section \(\approx 294 \text{ cm}^2\) (M1A1A1A1M1A1A1, N7).

Working:
Using Trigonometry
Evidence of using the cosine rule or sine rule (M1).
\( w^2 = 12^2 + 12^2 – 2 \times 12 \times 12 \times \cos 2.08 \) or \( \frac{w}{\sin 2.08} = \frac{12}{\sin 0.530796…} \) (A1).
\( w \approx 20.6977 \) or \( \frac{w}{2} \approx 10.3488 \) (A1).

Method 1: Sector and Triangle Approach
Geometric configuration: Points A, B, C, D, E, F, G, H as per diagram.

Segment AHB = sector OAB – triangle OAB (M1).
\( \frac{1}{2} \times 12^2 \times 2.08 – \frac{1}{2} \times 12^2 \times \sin 2.08 \approx 149.76 – 62.8655 \approx 86.8944 \) (A1).
Valid approach to find total cross-sectional area (M1).
Total area = \( \text{segment AHB} + \text{rectangle CDBA} \) (A1).
Rectangle CDBA width \( w \approx 20.6977 \), height 10 cm, area \( 10 \times 20.6977 \approx 206.977 \) (A1).
Total area \( 86.8944 + 206.977 \approx 293.871 \) (A1).
Method 2: Trapezium Approach
Trapezium CGOA = \( \frac{1}{2} \times (10 + (10 – 12 \cos 1.04)) \times \frac{20.6977}{2} \approx 72.0557 \) (A1).
Total area = \( 2 \times 72.0557 + \frac{1}{2} \times 12^2 \times 2.08 \approx 144.111 + 149.76 \approx 293.871 \) (A1).
Rounded: \( 293.871… \) or \( 294.430… \), final \( 294 \text{ cm}^2 \).

[7 marks]

(c) The gutter will overflow (180,000 cm³ > 176,658 cm³) (A1M1A1A1A1, N5).

Working:
Method 1: Volume of gutter = \( 294 \times 600 \approx 176,400 \text{ cm}^3 \) (adjusted to markscheme range 176,323 or 176,658) (A1).
Recognition that rainfall is an integral (M1).
\( R(t) = \int_0^{60} \left( 50 \cos\left(\frac{2\pi t}{5}\right) + 3000 \right) dt \).
Integrate: \( \frac{250}{2\pi} \sin\left(\frac{2\pi t}{5}\right) + 3000t \), evaluate from 0 to 60 (A1).
At \( t = 60 \): \( \sin(24\pi) = 0 \), result = \( 3000 \times 60 = 180,000 \text{ cm}^3 \) (A1).
Compare: \( 180,000 > 176,658 \), the gutter will overflow (A1).
Method 2: Minimum rate \( R'(t) \geq -50 + 3000 = 2950 \text{ cm}^3/\text{s} \), total \( 2950 \times 60 = 177,000 \text{ cm}^3 > 176,658 \), overflow.
Method 3: Time to fill \( 176,658 / 3000 \approx 58.886 \text{ s} < 60 \text{ s} \), overflow.

[5 marks]

Total [15 marks]