1.Пусть x меньшая сторона,тогда (х+4) большая
известно что s=140
х(х+4)=140
х^2+4х-140=0
D=4^2-4*1*(-140)=16+560+576
x1=-4+24/2=10
x2=-4-24/2=-14 -не удовлетворяет условие
меньшая сторона=10
большая 10+4=14
(9х^4у^2)(-8х^3у^6)=-72х^7у^8
Ответ:
![ctg(\alpha)=-\frac{7}{24}\\sin(\alpha)=-\frac{7}{25}\\cos( \alpha)=\frac{24}{25}](https://tex.z-dn.net/?f=ctg%28%5Calpha%29%3D-%5Cfrac%7B7%7D%7B24%7D%5C%5Csin%28%5Calpha%29%3D-%5Cfrac%7B7%7D%7B25%7D%5C%5Ccos%28+%5Calpha%29%3D%5Cfrac%7B24%7D%7B25%7D)
Объяснение:
(IV четверть, ctg, tg и sin отрицательные, cos положительный.
![sin^{2}( \alpha )+cos^{2}( \alpha )=1/cos^{2}( \alpha )\\tg^{2}( \alpha)+1=\frac{1}{cos^{2}( \alpha )} \\cos^{2}( \alpha)=\frac{1}{tg^{2}( \alpha)+1} =\frac{1}{\frac{49}{576}+1 } =\frac{576}{625} \\cos( \alpha)=\frac{24}{25} \\sin(\alpha)=-\frac{7}{25} \\ctg(\alpha)=-\frac{7}{24}](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28+%5Calpha+%29%2Bcos%5E%7B2%7D%28+%5Calpha+%29%3D1%2Fcos%5E%7B2%7D%28+%5Calpha+%29%5C%5Ctg%5E%7B2%7D%28+%5Calpha%29%2B1%3D%5Cfrac%7B1%7D%7Bcos%5E%7B2%7D%28+%5Calpha+%29%7D+%5C%5Ccos%5E%7B2%7D%28+%5Calpha%29%3D%5Cfrac%7B1%7D%7Btg%5E%7B2%7D%28+%5Calpha%29%2B1%7D+%3D%5Cfrac%7B1%7D%7B%5Cfrac%7B49%7D%7B576%7D%2B1+%7D+%3D%5Cfrac%7B576%7D%7B625%7D+%5C%5Ccos%28+%5Calpha%29%3D%5Cfrac%7B24%7D%7B25%7D+%5C%5Csin%28%5Calpha%29%3D-%5Cfrac%7B7%7D%7B25%7D+%5C%5Cctg%28%5Calpha%29%3D-%5Cfrac%7B7%7D%7B24%7D)