Attempting to teach my sister trig. I'm completely unsure how to approach the following question. She has learnt about the unit circle, sine and cosine laws.
I thought it was asking to determine how much the dog could reach so I modeled it as a circle enclosed in the triangle with a radius of 5m. Any advice on how to think about this question would be greatly appreciated.
I got $15.7$m by dividing the circumference of a theorized circle in half, but the answer is $15.2$m, is the top right portion of the circle being cut by the triangle $PBX$ (if it were drawn, where $XB>CB$)?
You're right that the answer is determined using a circle of radius $5$ meters, with the center at $P$, as shown below,
where $D$, $F$, $G$ and $H$ are the intersection points of the triangle with the circle. In particular, the dog can reach any of parts of the fence that are within $5$ meters of the leash's tether point at $P$, with this being those sections enclosed in the circle (i.e., the triangle lines shown in green).
Since $|PD| = |PF| = 5$, we have that $\triangle DPF$ is isosceles. Thus, the angle bisector of $\angle DPF$ meets $CB$ at $E$ at a right angle so, by ASA (angle-side-angle), we have $\triangle PED \cong \triangle PEF$. This means $|DE|=|EF|=a$ for some $a \gt 0$. Next, since $\measuredangle PEB = 90^{\circ}$, we then have from $\triangle PEB$ that
$$\sin(41^{\circ}) = \frac{|EP|}{|PB|} = \frac{h}{6.5} \; \; \to \; \; h = 6.5\sin(41^{\circ}) \approx 4.26438\,\text{m}$$
Using the Pythagorean theorem with $\triangle PEF$, we get
$$|PE|^2+|EF|^2=|PF|^2 \; \to \; h^2 + a^2 = 5^2 \; \to \; a = \sqrt{25 - h^2} \approx 2.6105\,\text{m}$$
As Ross Millikan's comment indicates, the total dog reachable length of the fence, $L$, would be the sum of the lengths of those displayed green triangle lines, i.e., the diameter $|GH| = 10\,\text{m}$, plus $|DF| = 2a$, to get that
$$L \approx 10 + 2(2.6105) \approx 15.2\,\text{m}$$