Kate crosses a road, of constant width 7 m, in order to take a photograph of a marathon runner, John, approaching at 3 m s⁻¹ - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 7

To find the minimum value of V, substitute the expression for V: $V = \frac{21}{24 \sin \theta + 7 \cos \theta}$ By substituting our earlier result: $V = \frac{21}{R} = \frac{21}{25} = 0.84$

Setting Kate's speed equal to V gives: $V = 0.84$ Thus, we substitute this into: $0.84 = \frac{21}{24 \sin \theta + 7 \cos \theta}$ From this, we can derive the equation for the distance.

To find the distance AB, we first express it using trigonometric identities: Using the sine rule, determine the distance using: $AB = \frac{7}{\sin \theta}$ We derived from part (b) that: $\sin \theta \approx \frac{7 AB}{175}$ Substituting values will yield AB. If simplified, we can calculate it numerically.

Using the value of Kate's speed to find θ: $1.68 = \frac{21}{24 \sin \theta + 7 \cos \theta}$ Rearranging gives: $24 \sin \theta + 7 \cos \theta = \frac{21}{1.68}$ Calculating this provides a relationship we can use to find θ.

We can solve for θ using: $\cos(\theta - \alpha) = 0.5$ This results in: $\theta - 73.74° = 60° \text{ or } \theta - 73.74° = -60°$ Thus, solving:

1. $\theta = 60° + 73.74° \approx 133.74°$
2. $\theta = -60° + 73.74° ext{ gives a value outside bounds}$ Final possible value: $\theta = 13.74°$. Hence the two possible values for θ are approximately 13.74° and 133.74°.