Step-by-step explanation:
c. ultraviolet energy
Hey!
So the first thing we realize is that it says that the equation is perpendicular to the line, meaning that the slope of the line is the negative reciprocal of the slope of the line you are given. Since we are given the slope of this line as 3/4 we can take the negative reciprocal of this to get -(4/3).
Now that we have the slope and a point on the line you can plug those into the equation y = mx + b to find b. The slope of the line is m and the point contains the x and y values.
5 = -(4/3)(-3) + b
5 = 4 + b
1 = b
Since we have the y-intercept and the slope now we can plug that into the slope-intercept form equation to get the equation we need:
y = -(4/3)x + 1
well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.
![\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5](https://tex.z-dn.net/?f=%5Cbf%20cos%28A%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B3%7D%7D%7B%5Cunderset%7Bhypotenuse%7D%7B5%7D%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bgetting%20the%20opposite%20side%7D%7D%7Bb%3D%5Cpm%5Csqrt%7B5%5E2-3%5E2%7D%7D%5Cimplies%20b%20%3D%20%5Cpm%204%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7BIV~Quadrant%7D%7Bb%20%3D%20-4%7D%5Cqquad%20%5Cqquad%20sin%28A%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-4%7D%7D%7B%5Cunderset%7Bhypotenuse%7D%7B5%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20cos%28B%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B12%7D%7D%7B%5Cunderset%7Bhypotenuse%7D%7B13%7D%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bgetting%20the%20opposite%20side%7D%7D%7Bb%3D%5Cpm%5Csqrt%7B13%5E2-12%5E2%7D%7D%5Cimplies%20b%20%3D%20%5Cpm%205)
![\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7BIV~Quadrant%7D%7Bb%20%3D%20-5%7D%5Cqquad%20%5Cqquad%20sin%28B%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-5%7D%7D%7B%5Cunderset%7Bhypotenuse%7D%7B13%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20sin%28A-B%29%3D%5Ccfrac%7B-4%7D%7B5%7D%5Ccdot%20%5Ccfrac%7B12%7D%7B13%7D-%5Cleft%28%20%5Ccfrac%7B3%7D%7B5%7D%5Ccdot%20%5Ccfrac%7B-5%7D%7B13%7D%20%5Cright%29%5Cimplies%20sin%28A-B%29%3D%5Ccfrac%7B-48%7D%7B65%7D%20-%20%5Cleft%28%20%5Ccfrac%7B-15%7D%7B65%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20sin%28A-B%29%3D%5Ccfrac%7B-48%7D%7B65%7D%20%2B%20%5Ccfrac%7B15%7D%7B65%7D%5Cimplies%20sin%28A-B%29%3D%5Ccfrac%7B-33%7D%7B65%7D)
You have to use the quadratic formula to solve this. The zeros of this quadratic are 2 + sqrt2/2 and 2 - sqrt2/2, which in "real" numbers is 1.707 and .2928
