Hello from MrBillDoesMath!
Answer:
sqrt(15)/4
Discussion:
The attachment shows a right triangle where arcsin(@) = 1/4. (The arcsin means find the angle whose sine is 1/4 as is done in the triangle.) The missing side, sqrt(15), shown in red, was determined by the Pythagorean theorem.
Then cos(@) = cos(arcsin(1/4)) = sqrt(15)/4
Thank you,
MrB
ANSWER
1a)
The given system of equation is:
![3x - 6y = - 12...(1)](https://tex.z-dn.net/?f=3x%20-%206y%20%3D%20%20-%2012...%281%29)
and
![x - 2y = - 8...(2)](https://tex.z-dn.net/?f=x%20-%202y%20%3D%20%20-%208...%282%29)
We make x the subject in equation (2) to get;
![x = 2y - 8...(3)](https://tex.z-dn.net/?f=x%20%3D%202y%20-%208...%283%29)
We put equation (3) into equation (1) to get;
![3(2y - 8) - 6y = - 12](https://tex.z-dn.net/?f=3%282y%20-%208%29%20-%206y%20%3D%20%20-%2012)
We expand the bracket to get,
![6y - 24 - 6y = - 12](https://tex.z-dn.net/?f=6y%20-%2024%20-%206y%20%3D%20%20-%2012)
![6y - 6y = 24 - 12](https://tex.z-dn.net/?f=6y%20-%206y%20%3D%2024%20-%2012)
This implies that,
![0 = 12](https://tex.z-dn.net/?f=0%20%3D%2012)
This statement is false.
Hence the system has no solution.
1b)
Let us write the two systems in the slope-intercept form;
For the first equation, we have
![3x - 6y = - 12](https://tex.z-dn.net/?f=3x%20-%206y%20%3D%20%20-%2012)
Thus implies that,
![- 6y = - 3x- 12](https://tex.z-dn.net/?f=%20-%206y%20%3D%20-%203x-%2012%20)
![y = \frac{1}{2} x + 2](https://tex.z-dn.net/?f=%20y%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20x%20%2B%202)
For the second equation, we have:
![- 2y = - x - 8](https://tex.z-dn.net/?f=%20-%20%202y%20%3D%20%20-%20x%20-%208)
![y = \frac{1}{2} x + 8](https://tex.z-dn.net/?f=%20y%20%3D%20%20%20%5Cfrac%7B1%7D%7B2%7D%20x%20%20%2B%208)
We can see that, the two equations are parallel since they have the same slope.
Also the two equations will never meet since they have different y-intercepts.
Hence the system has no solution.