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 2 - 2013 - Paper 7

To find the minimum value of V, we analyze the formula:

$V = \frac{21}{24\sin{θ} + 7\cos{θ}}$

To minimize V, we maximize the denominator. From the earlier expression, we can use calculus or optimization techniques, yielding a maximum value of the form

$\max(24\sin{θ} + 7\cos{θ}) = 21$

Thus the minimum value of V becomes:

$V_{min} = 0.84$

Using the trigonometric identity, we apply the following relation:

$θ = \arcsin{\left( \frac{7}{AB} \right)}$

Hence, we rearrange for AB and substitute:

$AB = \frac{7}{\sin{θ}}$

Approximating further leads to:

$AB \approx 7.29 m$

We start with the formula for V:

$1.68 = \frac{21}{24\sin{θ} + 7\cos{θ}}$

Rearranging gives:

$24\sin{θ} + 7\cos{θ} = \frac{21}{1.68}$

Now evaluate the left-hand side:

$Rcos(θ - α) = \frac{21}{1.68}$

Solving this results in two possible angles for θ, called θ₁ and θ₂, yielding:

$θ_1 ≈ 60° \quad and \quad θ_2 ≈ 117°$