14. Two ferries leave Tsawwassen, BC, at 1:00 p.m. The first vessel travels on a bearing of $\mathrm{N} 68^{\circ} \mathrm{W}$ toward Departure Bay at a speed of $26 \mathrm{~km} / \mathrm{h}$. The second vessel travels at a speed of $35 \mathrm{~km} / \mathrm{h}$ on a bearing of $\mathrm{S} 33^{\circ} \mathrm{W}$ toward Swartz Bay. How far apart are the ferries at $1: 30$ p.m.? A $19.5 \mathrm{~km}$ B $23.3 \mathrm{~km}$ C $23.7 \mathrm{~km}$ D $47.4 \mathrm{~km}$

Step 1 ：Label the starting point of the first vessel as A and the starting point of the second vessel as B. Let C be the point where the first vessel is at 1:30 p.m., and D be the point where the second vessel is at 1:30 p.m.

Step 2 ：Since the first vessel travels at a speed of 26 km/h, it travels a distance of 26 \(\frac{1}{2}\) = 13 km in 30 minutes. Similarly, the second vessel travels a distance of 35 \(\frac{1}{2}\) = 17.5 km in 30 minutes.

Step 3 ：The angle between the paths of the two vessels is 68 + 33 = 101 degrees.

Step 4 ：Using the law of cosines, we can find the distance between C and D: \(CD^2 = 13^2 + 17.5^2 - 2 \cdot 13 \cdot 17.5 \cdot \cos 101^\circ\)

Step 5 ：Calculate the value of CD^2: \(CD^2 = 169 + 306.25 - 2 \cdot 13 \cdot 17.5 \cdot (-0.1564)\approx 543.3\)

Step 6 ：Take the square root of CD^2 to find the distance between the ferries at 1:30 p.m.: \(CD = \sqrt{543.3} \approx \boxed{23.3}\) km