How to calculate distance of shortest path when there is a flow in the river? I am trying to solve this question:


I have figured out a solution involving calculus, however there seems to be a much simpler solution that was provided with the question, as follows:

However, I can't for the life of me understand this solution? Can someone explain what is happening in this solution?
It appears that a distance diagram is being used: the top of the triangle represents the distance traveled by the flow of the river in time t (the velocity of the river is 4 m/s); the left side of the triangle represents the distance traveled by the boat in time t (the velocity of the boat is 1 m/s).
This makes sense, however I can't understand the right side of the triangle, which represents the distance of the shortest path. How does this third side of the triangle represent the shortest distance? This is the part that is confusing me. If I can understand that, the rest of the solution is fairly straightforward.
 A: The river is flowing from left to right, $\ t\ $ is the time the woman takes to go from one bank to the other, and hence the total distance she rows relative to the moving stream, and $\ \alpha\ $ is the angle which the direction she's pointing the boat makes with the opposite bank.
During the time she takes to cross the river, it will carry her a distance $\ 4t\ $ downstream to the apex of the angle $\ \phi\ $.  Thus, since the woman departs from the apex of the angle $\ \beta\ $, and arrives at the apex of the angle $\ \phi\ $, then the distance she has travelled relative to the land is $\ d\ $.
Since $\ d=\frac{60}{\sin\phi} $, then $\ d\ $ will be a minimum when $\ \sin\phi\ $ is a maximum, but since $\ \sin\phi=\frac{\sin\beta}{4}\ $, then $\ \sin\phi\ $ cannot exceed $\ \frac{1}{4}\ $, and it does have this value when $\ \beta=90^{\large\circ}\ $.
A: The Triangle Law of Vector Addition states that when two vectors are represented by two sides of a triangle in the same order then the third side of that triangle represents the resultant of the vectors. Note that the two vctors to be added must be placed head to tail.
Let  $ \vec a $  denote the velocity of the boat in still water and   $ \vec b $  denote the velocity of the water. If  $ \vec v $ denotes the actual velocity of the boat, then $\vec v= \vec a+ \vec b$.
Suppose that the boat crosses the river by the shortest path in a time $t$. After a time $t$, let $ \vec x$ denote the displacement of the boat in still water and $ \vec y$ denote the displacement of the water. At this time, let  $ \vec r$ denote the actual displacement of the boat due to the combined effect of the effort of the rower and the current in the river. Then
$$\vec r=\vec x+ \vec y$$.
By the Triangle Law, $ \vec x,\ \vec y\  and\  \vec  r $ may be represented by the sides of a triangle.
$\vec y$ is easy to draw as it has magnitude $4t$ and is parallel to the bank.
Assuming that the river is flowing left to right, $\vec x$ must be in some random direction up the river but is of magnitude $1t$. By the Triangle Law, $\vec x$ and $\vec y$ must be drawn head to tail. The resultant displacement  $\vec r$ must now be drawn from the starting point to the head of $\vec y$ and is of magnitude d, the actual distance travelled by the boat.
As the questioner has pointed out, the problem is straightforward after the triangle is correctly drawn and labelled.
As $d =\frac{60}{sin\phi}$, d is a minimum when $sin \phi$ is a maximum. Then $sin \phi = \frac{sin \beta}{4}$ yields $( sin \phi) _{max} = \frac{1}{4}$.
So $\frac {60}{d_{min}}= \frac{1}{4}$ gives $d_{min}= 240m$.
When applying the Triangle Law, note that if the two vectors to be added are cyclic, then the resultant must be anti-cyclic.
