Conor's property is bounded by the straight bank of a river, as shown in Figure 1 above - Leaving Cert Mathematics - Question 9 - 2017

To show this, we rely on the second part of the triangle: Since angle $E$ is $30^{\text{o}}$, we use the relationship: $tan(30^{\text{o}}) = \frac{|E|}{|DT|}$ This leads to: $|E| = |DT| \cdot tan(30^{\text{o}}) = \frac{|DT|}{\sqrt{3}}$ Now, using the Pythagorean theorem on triangle $ACT$: $|CT|^2 + 15^2 = |DT|^2$ Substituting $|DT|$ into $|E|$, allows us to prove: $\sqrt{3}|CT| = \sqrt{225+|CT|^2}{3}$

From the previous steps, we have: $\sqrt{3}|CT| = \sqrt{225 + |CT|^2}$ Squaring both sides gives: $3|CT|^2 = 225 + |CT|^2$ Simplifying leads to: $2|CT|^2 = 225$ Thus: $|CT|^2 = 112.5$ And taking the square root provides: $|CT| = \sqrt{112.5} \approx 5.3\, m$

Now we substitute $|CT|$ back: $|E| = \sqrt{3} |CT| = \sqrt{3} \cdot 5.3 \approx 9.2 \text{ m}$

Using the cosine rule, we have based on previous angles: $cos(\theta) = \frac{|CT|}{|FT|}$ Using our known values results in: $\theta \approx 54.7^{\text{o}}$

If the tree can fall into any direction, the probability can be given by considering the relevant angle: $P = \frac{(54.7)(2)}{360} \approx 0.3038$ Thus as a percentage: P \approx 30.4 \text{ %}