2020 H2 Maths Paper 1, Question 11

Since $AB\bot XY$, triangles $\text{AKC}$ and $\text{AKD}$ are right-angled at $A$. Let $\alpha = \angle\text{AKC}$ and $\beta = \angle\text{AKD}$. With $AC = a$ and $AK = x$,

Since $AD = AC + CD = a + 4$,

From the diagram, $\theta = \angle\text{CKD} = \beta - \alpha$. Using the tangent subtraction formula,

Multiplying numerator and denominator by $x^{2}$,

Treating $a$ as constant, we differentiate $\tan\theta = \frac{4x}{x^{2} + a^{2} + 4a}$ with respect to $x$ by the quotient rule, writing $c = a^{2} + 4a$ for brevity.

Setting this to zero for a stationary value,

taking the positive root since $x$ is a length. Substituting back,

The optimal point gives the largest angle $\theta$, but the corresponding kicking distance $AK = x = \sqrt{a(a + 4)}$ may be too far from the posts for the player to strike the ball with enough power or accuracy.

Triangle $\text{AKD}$ is right-angled at $A$, so

At the optimal point, $x = \sqrt{a(a + 4)}$, so

When $a$ is much greater than $4$, $\frac{a}{a + 4} \rightarrow 1$, so $\tan\left( \angle\text{KDA} \right) \rightarrow 1$.

Since $XC = DY$ and $XY = XC + CD + DY = 50$,

As $A$ varies between $X$ and $C$, $a = AC$ ranges strictly between $0$ (at $A = C$) and $23$ (at $A = X$).

From part (ii), the optimal angle satisfies $\tan\theta = \frac{2}{\sqrt{a(a + 4)}}$, which decreases as $a$ increases (since $a(a + 4)$ increases). So $\theta$ is largest when $a$ is smallest and smallest when $a$ is largest.

As $a \rightarrow 0^{\text{+}}$ (i.e. $A \rightarrow C$), $\tan\theta \rightarrow \infty$, so $\theta \rightarrow 90^{\circ}$.

As $a \rightarrow 23^{-}$ (i.e. $A \rightarrow X$),

giving $\theta \approx {4.59}^{\circ}$.