Помогите срочно нужно решит 8 9 10
AB=7
AB=7?
da
да
угол там есть 8 задаче АСВ ?
нет
а NC что биссектриса или что
точно?
По теореме косинусов найдем угол АВС он будет равен ACB=arccos(1/5) решая систему получим AN=35/11 2) не видно ! 3)по теореме синусов BC/sin135 = AC/sinABC BC=2√2*sin135/sin30=4
![AN^2=NC^2+5^2-2*NC*5*cos(\frac{arccos(1/5)}{2})\\
BN^2=NC^2+6^2-2*NC*6*cos(\frac{arccos(1/5)}{2})\\
AN+BN=7\\
\\
AN=7-BN\\\
\\
(7-BN)^2=NC^2+25-10*NC*cos(\frac{arccos(1/5)}{2})\\
BN^2=NC^2+36-12NC*cos(\frac{arccos(1/5)}{2})\\
\\
24-14BN+BN^2=NC^2-10NC*cos(\frac{arccos(1/5)}{2})\\
BN^2=NC^2+36-12BN*cos(\frac{arccos(1/5)}{2})\\
\\
cos(\frac{arccos(1/5)}{2})=\sqrt{\frac{3}{5}}](https://tex.z-dn.net/?f=AN%5E2%3DNC%5E2%2B5%5E2-2%2ANC%2A5%2Acos%28%5Cfrac%7Barccos%281%2F5%29%7D%7B2%7D%29%5C%5C%0ABN%5E2%3DNC%5E2%2B6%5E2-2%2ANC%2A6%2Acos%28%5Cfrac%7Barccos%281%2F5%29%7D%7B2%7D%29%5C%5C%0AAN%2BBN%3D7%5C%5C%0A%5C%5C%0AAN%3D7-BN%5C%5C%5C%0A%5C%5C%0A%287-BN%29%5E2%3DNC%5E2%2B25-10%2ANC%2Acos%28%5Cfrac%7Barccos%281%2F5%29%7D%7B2%7D%29%5C%5C%0ABN%5E2%3DNC%5E2%2B36-12NC%2Acos%28%5Cfrac%7Barccos%281%2F5%29%7D%7B2%7D%29%5C%5C%0A%5C%5C%0A24-14BN%2BBN%5E2%3DNC%5E2-10NC%2Acos%28%5Cfrac%7Barccos%281%2F5%29%7D%7B2%7D%29%5C%5C%0ABN%5E2%3DNC%5E2%2B36-12BN%2Acos%28%5Cfrac%7Barccos%281%2F5%29%7D%7B2%7D%29%5C%5C%0A%5C%5C%0Acos%28%5Cfrac%7Barccos%281%2F5%29%7D%7B2%7D%29%3D%5Csqrt%7B%5Cfrac%7B3%7D%7B5%7D%7D "AN^2=NC^2+5^2-2*NC*5*cos(\frac{arccos(1/5)}{2})\\
BN^2=NC^2+6^2-2*NC*6*cos(\frac{arccos(1/5)}{2})\\
AN+BN=7\\
\\
AN=7-BN\\\
\\
(7-BN)^2=NC^2+25-10*NC*cos(\frac{arccos(1/5)}{2})\\
BN^2=NC^2+36-12NC*cos(\frac{arccos(1/5)}{2})\\
\\
24-14BN+BN^2=NC^2-10NC*cos(\frac{arccos(1/5)}{2})\\
BN^2=NC^2+36-12BN*cos(\frac{arccos(1/5)}{2})\\
\\
cos(\frac{arccos(1/5)}{2})=\sqrt{\frac{3}{5}}")