I reference the website below, for my two questions. http://www.delphiforfun.org/Programs/Math_Topics/WindTriangle.htm
References:
Plane is heading north
Wind is to the SE
a = ground speed vector
b = Wind Vector = 25
c = Air speed with heading Vector = 31
A = angle between B and C
B = "Wind Correction Angle"
C = "Wind to Track Angle"
1) I understand everything in this picture except for why I have to use 135$^{\circ}$ for the "Wind to Track Angle" (C). I know the website says to just subtract Desired Course - Wind direction but obviously 0$^{\circ}$ - 45$^{\circ}$ is -45$^{\circ}$. If I use -45$^{\circ}$ the math works out to have the Wind Correction angle as -34.8$^{\circ}$ and 43.1 ground speed, which is not correct. I'm also getting confused when I should use a negative angle or not. An example is in the triangle 2 picture where the wind is towards the east.
2) I understand the concept of splitting the airspeed vector to a right angle to come up with the ground speed. My question is how can I find the ground speed using the Law of Sines if I wanted to? My geometry is extremely rusty but I remember that a triangle's angles add up to 180$^{\circ}$. If that is the case then 135$^{\circ}$ + 34.8$^{\circ}$ + A = 180$^{\circ}$ . Therefore, A = 10.2$^{\circ}$ . What happens also if the "Wind to Track Angle" is negative?
Now I thought I could use the law of Sines and come up with the equation
a = 25 * sin(10.2) / sin(34.8)
a = 7.757163
The answer is very close but if you do the "split method" as descirbed on the website you get a = 7.787993
Sorry for the bad math formatting, not very good at LATEX. Please let me know if you need any clarifications. If you want to play around with different wind speed directions the .exe on the website above works well and is where I got the linked picture from.
The reason the angle is $135$ degrees rather than $-45$ is that wind direction is traditionally given as the direction the wind is blowing from. The wind vector, on the other hand, points in the direction the wind is blowing toward. For that reason, the instructions tell you to add $180$ degrees to the "wind direction" given in the problem.
So a wind whose direction is given as $45$ degrees (northeast) is said to be blowing to the southeast, which is $-135$ degrees.
Once you have the correct angle of the wind vector, the law of sines should work as you suggest. Just be aware that you are measuring your angles with just a few digits of precision, and this can affect the precision of your final answer. In particular, the angle given as $10.2$ degrees could be anything in the range $10.15$ to $10.24999$--anybody those numbers rounds to $10.2.$ So you can have almost $1/2$ percent error in this value. Since it is a smallish angle, the percent error of its sine is almost as large, and using the law of sines, that error is transmitted directly to the answer (in addition to the error caused by the rounding of the other angle). The difference you found between the two answers is less than $1/2$ percent, so they are completely consistent with each other.