Якщо точка лежить на колі, то
1.Безліч
2.Один
3.Один
Найдем угол между векторами AB и АС, для этого для начала определим координаты вектора АВ и вектора АС.
![\overline{AB}=\{4-0;6-6\}=\{4;0\}\\ \overline{AC}=\{3\sqrt{3}-0;3-6\}=\{3\sqrt{3};-3\}](https://tex.z-dn.net/?f=%5Coverline%7BAB%7D%3D%5C%7B4-0%3B6-6%5C%7D%3D%5C%7B4%3B0%5C%7D%5C%5C%20%5Coverline%7BAC%7D%3D%5C%7B3%5Csqrt%7B3%7D-0%3B3-6%5C%7D%3D%5C%7B3%5Csqrt%7B3%7D%3B-3%5C%7D)
Косинус угла между векторами AB и AC:
![\cos \angle BAC =\dfrac{\overline{AB}\cdot \overline{AC}}{|\overline{AB}|\cdot |\overline{AC}|}=\dfrac{4\cdot 3\sqrt{3}+0\cdot (-3)}{\sqrt{4^2+0^2}\cdot \sqrt{(3\sqrt{3})^2+(-3)^2}}=\dfrac{4\cdot3\sqrt{3}}{4\cdot 6}=\dfrac{\sqrt{3}}{2}\\ \\ \\ \angle BAC=\arccos\Bigg(\dfrac{\sqrt{3}}{2}\Bigg)=\dfrac{\pi}{6}=30^\circ](https://tex.z-dn.net/?f=%5Ccos%20%5Cangle%20BAC%20%3D%5Cdfrac%7B%5Coverline%7BAB%7D%5Ccdot%20%5Coverline%7BAC%7D%7D%7B%7C%5Coverline%7BAB%7D%7C%5Ccdot%20%7C%5Coverline%7BAC%7D%7C%7D%3D%5Cdfrac%7B4%5Ccdot%203%5Csqrt%7B3%7D%2B0%5Ccdot%20%28-3%29%7D%7B%5Csqrt%7B4%5E2%2B0%5E2%7D%5Ccdot%20%5Csqrt%7B%283%5Csqrt%7B3%7D%29%5E2%2B%28-3%29%5E2%7D%7D%3D%5Cdfrac%7B4%5Ccdot3%5Csqrt%7B3%7D%7D%7B4%5Ccdot%206%7D%3D%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5C%5C%20%5C%5C%20%5C%5C%20%5Cangle%20BAC%3D%5Carccos%5CBigg%28%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5CBigg%29%3D%5Cdfrac%7B%5Cpi%7D%7B6%7D%3D30%5E%5Ccirc)
Ответ: 30°.
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3,8-1,5y+(4,5y-0,8)=2,4y+3
3,8-1,5y+4,5y-0,8=2,4y+3
-1,5y+4,5y-2,4y =-3,8+0,8+3
0,6у=0
у=0