Current Attempt in Progress
A sailboat race course consists of four legs, defined by the displacement vectors $\vec{A}, \vec{B}, \vec{C}$, and $\vec{D}$, as the drawing indicates. The magnitudes of the first three vectors are $A=3.50 \mathrm{~km}, B=4.60 \mathrm{~km}$, and $C=5.10 \mathrm{~km}$. The finish line of the course coincides with the starting line. Using the data in the drawing, find (a) the distance of the fourth leg and (b) the angle $\theta$.
(a) Number
(b) Number
i
i
Units
Units

Step 1 ：Since the finish line coincides with the starting line, we have $\vec{A} + \vec{B} + \vec{C} + \vec{D} = \vec{0}$

Step 2 ：Thus, $\vec{D} = - (\vec{A} + \vec{B} + \vec{C})$

Step 3 ：Let $\alpha$ be the angle between $\vec{A}$ and $\vec{B}$, and $\beta$ be the angle between $\vec{B}$ and $\vec{C}$

Step 4 ：Using the Law of Cosines, we have $D^2 = A^2 + B^2 - 2AB \cos \alpha + B^2 + C^2 - 2BC \cos \beta + A^2 + C^2 - 2AC \cos (\alpha + \beta)$

Step 5 ：Substitute the given values, $D^2 = 3.50^2 + 4.60^2 - 2(3.50)(4.60) \cos 120^\circ + 4.60^2 + 5.10^2 - 2(4.60)(5.10) \cos 90^\circ + 3.50^2 + 5.10^2 - 2(3.50)(5.10) \cos 150^\circ$

Step 6 ：Calculate the value of $D^2$, $D^2 = 3.50^2 + 4.60^2 - 2(3.50)(4.60) \cos 120^\circ + 4.60^2 + 5.10^2 + 3.50^2 + 5.10^2 - 2(3.50)(5.10) \cos 150^\circ$

Step 7 ：Calculate the value of $D$, $D = \sqrt{D^2} = \boxed{4.20}$

Step 8 ：Using the Law of Cosines to find $\theta$, we have $D^2 = A^2 + C^2 - 2AC \cos \theta$

Step 9 ：Substitute the given values, $4.20^2 = 3.50^2 + 5.10^2 - 2(3.50)(5.10) \cos \theta$

Step 10 ：Solve for $\cos \theta$, $\cos \theta = \frac{3.50^2 + 5.10^2 - 4.20^2}{2(3.50)(5.10)}$

Step 11 ：Calculate the value of $\cos \theta$, $\cos \theta = \frac{3.50^2 + 5.10^2 - 4.20^2}{2(3.50)(5.10)}$

Step 12 ：Find the value of $\theta$, $\theta = \arccos \left( \frac{3.50^2 + 5.10^2 - 4.20^2}{2(3.50)(5.10)} \right) = \boxed{75.0^\circ}$