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)
The Answer: 3/12, 5/6, 4/5...
Where's the "triangle with alt. BD?" This problem can be solved without the diagram, but the solution would be easier with it.
BD is the altitude. Find the length of BD by finding the dist. between (-1,4) and (2,4); it is 2-(-1), or 3. |BD| = 3.
I've graphed the triangle myself and have found that the "base" of the triangle is the vertical line thru (2,1) and (2,6); its length is 6-1, or 5.
Thus, the area of this triangle is A = (b)(h) / 2, or A = (5)(3) / 2 = 10/3 square inches.
