Joan is playing golf - Leaving Cert Mathematics - Question 9 - 2015

Using the Law of Cosines:

$|AH|^2 = |AT|^2 + |TH|^2 - 2|AT||TH|\cos(18^{\circ})$

Substituting the given values:

$|AH|^2 = 190^2 + 385^2 - 2(190)(385)\cos(18^{\circ})$

Calculating the distance gives:

$|AH| \approx 213 \text{ m}$

The formula for the height of the ball is given as:

$h = -6t^{2} + 22t + 8$

Substituting $t = 0$ (at the point of hitting) gives:

$h = 8 ext{ m}$

From the height calculated, we have:

$|OB| = 38t$

We know from the previous calculations that when $t = 4$, $|OB| = 152$ m. Thus:

$\tan(\angle ZBK) = \frac{8}{152}$

Calculating gives:

$\angle ZBK \approx 3^{\circ}$

From the geometric relationships:

$d = \frac{h}{1} - \frac{25}{2}$

Thus,

$|CD| = 25 - h$.

Using the relationships established, we can set up:

$d^{2} + |CD|^{2} = 25^{2}$

Substituting for $d$ and $|CD|$:

$(2h)^{2} + (25 - h)^{2} = 625$

Solving the resulting equation leads to:

$h = 10 ext{ m}$