(a)
Vertical height difference: \( 3.1 – 1.0 = 2.1 \) m
Use sine ratio: \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2.1}{2.8} = 0.75 \) (M1)
\( \theta = \arcsin(0.75) \approx 48.6^\circ \) (48.5903…°) (A1)
Result: \( 48.6^\circ \) [2]

(b)
METHOD 1
Find horizontal distance from \( C_1 \) to \( R \): \( \sqrt{2.8^2 – 2.1^2} = \sqrt{7.84 – 4.41} = \sqrt{3.43} \approx 1.85 \) m (M1)(A1)
Horizontal distance from \( C_2 \) to \( R \): \( 6.4 – 1.85 \approx 4.55 \) m (A1)
Distance \( C_2 \) to \( R \): \( \sqrt{4.55^2 + 2.1^2} \approx \sqrt{20.7025 + 4.41} \approx \sqrt{25.1125} \approx 5.01 \) m (A1)
METHOD 2
Use cosine rule with angle from part (a): \( c^2 = 2.8^2 + 6.4^2 – 2 \times 2.8 \times 6.4 \cos(48.6^\circ) \) (M1)(A1)
\( c^2 \approx 7.84 + 40.96 – 35.84 \times 0.6691 \approx 48.8 – 23.984 \approx 25.1125 \)
\( c \approx \sqrt{25.1125} \approx 5.01 \) m (A1)(A1)
Result: \( 5.01 \) m [4]

(c)
Camera 1 is closer to the cash register (2.8 m) than Camera 2 (5.01 m), and both cameras are at the same height (3.1 m) (R1)
For the same vertical drop (2.1 m), a shorter distance gives a larger angle of depression, as \( \sin \theta = \frac{2.1}{\text{distance}} \), so \( \theta_1 > \theta_2 \) (A1)
Result: Camera 1 [2]