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 for this question is 4.
Answer:
<em>after 4seconds</em>
Step-by-step explanation:
Given the height, h , in feet, of the football above the ground after t seconds expressed by h ( t ) = − 8 t^2 + 32 t, the height of the ball on the ground is 0feet.
Substitute h(t) = 0 into the expression and calculate t;
h ( t ) = − 8 t^2 + 32 t
0 = − 8 t^2 + 32 t
8t² = 32t
8t = 32
Divide both sides by 8
8t/8 = 32/8
<em>t = 4s</em>
<em>Hence the football hits the ground after 4seconds</em>
sin(x) = 2cos(x)
tan(x) = 2
tan⁻¹[tan(x)] = tan⁻¹(2)
x ≈ 63.4
sin(2x) = sin[2(63.4)]
sin(2x) = sin(126.8)
sin(2x) ≈ 0.801
So the tower is going to be larger than the shadow but not to large so let work from there and not to large so the only logical answer is d
