OK - First thing to say is you have done well. The components you have are the same as I have .. in fact, you pretty well have all the answers you need. Also, since you have put in the effort, you get the explanation. There are places where your understanding is nearly there, but not quite.
So try and stay with this all the way to the end, because on the way, in addition to getting the numerical answers, we want that you just do not fear vector sums of any type.
So now, we can talk you through this, setting out what is going on. This is a problem of
two separate vector additions, where the resultant of the first is used as the starting point for the second.
The first vector we can call \,\vec{240wow} is 240 long at an angle of 0 degrees.
For the graphical diagram method, you can draw a vector 240mm long. (You need a BIG piece of paper). There was a time it would be very big, with very thin sharp lines, but now, we only need a visual aid, and we get to the distances using a calculator. Left to right, you are already assuming x-axis represents W-E, and y-axis represents S-N. That is OK! Probably you set its origin at the starting point.
Now at the head of \,\vec{240wow} you start the foot of the second vector \,\vec{96yay}.
To help you draw it, you put a little 'x' and 'y' axes drawn in lightly with its new local origin at the start of \,\vec{96yay}.
You know it is 96 long, and then draw it with the angle 28 degrees N of E, so rotate anticlockwise by 28 degrees.
At this point fix in your mind that you have decided anticlockwise angles mean positive angles !.
As you have successfully calculated, the head of \,\vec{96yay} ends up 87.76m further to the EAST, and 45.07m to the North.
So the
resultant of adding \,\vec{240wow} to \,\vec{96yay} is the vector from the starting point to the ending point of the second hit. He
could have hit it there in the first place, but instead, it took two drives. You
could have used your second equation c^2 = a^2 + b^2 -2abcosC to get the length, but we are going to duck that bit,
because we already have the x-y components, so we can choose not to fall over ourselves doing what we don't need to.
Notice that the x_component (E-W) of the first drive was 240. The y_component (N-S) was zero.
x240=\vec{240wow}\cdot\,cos(0)\,\,=\,\,240
y240=\vec{240wow}\cdot\,sin(0)\,\,=\,\,0
For the second drive, you have calculated the answers - but this is how they fit.
x96=\vec{96yay}\cdot\,cos(28)\,\,=\,\,87.76
y96=\vec{96yay}\cdot\,sin(28)\,\,=\,\,45.07
Now we can add a third vector. It can be seen as adding onto
the resultant of the first two, or simply all three adding together heads-to-tails. The key thing here is vector addition is what is known as
associative, which is a geek word meaning.. "you can add them up grouped any which way you like". If 3 vectors were called u, v, and w
u + (v + w) = (u + v) + w ... easy enough.
Now we add in the final vector, \,\vec{12duh}
As before, at its start point, it gets a little local x-y axes, so we can point its direction 25 degrees S of East. . The angle you need can be thought of in two ways. It was a
positive angle of 335 degrees (remember going anti-clockwise), or it was -25 degrees (going clockwise). Either way will work!
x12=\vec{12duh}\cdot\,cos(-25)\,\,=\,\,\vec{12duh}\cdot\,cos(335)\,\,=\,\,10.87 , which was an answer you got.
y12=\vec{12duh}\cdot\,sin(-25)\,\,=\,\,\vec{12duh}\cdot\,sin(335)\,\,=\,\,-5.07 , which by now you can see means it gets back closer to the line of the first hit.
The last part teaches you that once all the vectors are resolved into their x and y components, you can add and subtract the components, and still end up with the right vector.
All the x-components were found by [magnitude] x cos(angle)
All the y-components were found by [magnitude] x sin(angle)
So we come to the hole-in-one.
Add up all the x-components, and get 338.64
Add up all the y-components, and get 39.997 (OK then make it 40)
Go on then. The grand final vector \,\vec{big.hit} will have a magnitude that had better be (a bit) greater than 340.
Pythagoras!
If you know all sides of that big final right triangle, you can probably figure the angle in it.
You should then have the [magnitude] and [direction] of the hole-in-one hit.
