Найдите острые углы прямоугольного треугольника, если его гипотенуза = 12 см, а площадь = 18 см²
Ответы
Ответ дал:
0
tg β = b/a
где![a= \frac{c+2r+ \sqrt{c^2-4cr-4r^2} }{2} ;b= \frac{c+2r- \sqrt{c^2-4cr-4r^2} }{2} a= \frac{c+2r+ \sqrt{c^2-4cr-4r^2} }{2} ;b= \frac{c+2r- \sqrt{c^2-4cr-4r^2} }{2}](https://tex.z-dn.net/?f=a%3D+%5Cfrac%7Bc%2B2r%2B+%5Csqrt%7Bc%5E2-4cr-4r%5E2%7D+%7D%7B2%7D+%3Bb%3D+%5Cfrac%7Bc%2B2r-+%5Csqrt%7Bc%5E2-4cr-4r%5E2%7D+%7D%7B2%7D+)
![P=a+b+c\\ P= \sqrt{(a+b)^2}+c \\ P= \sqrt{a^2+2ab+b^2}= \sqrt{c^2+4S}+c= \sqrt{12^2+4\times18}+12=6 \sqrt{6} +12\\ r= \frac{2S}{P}= \frac{2\cdot 18}{6 \sqrt{6} +12}= \frac{2\cdot 18(6\sqrt{6}-12)}{(6\sqrt{6})^2-12^2} =3\sqrt{6}-6\\ a= \frac{c+2r+ \sqrt{c^2-4cr-4r^2} }{2}= \frac{12+2(3\sqrt{6}-6)- \sqrt{12^2-4\cdot 12(3\sqrt{6}-6)-4(3\sqrt{6}-6)^2} }{2}=3\sqrt{6}+3\sqrt{2} P=a+b+c\\ P= \sqrt{(a+b)^2}+c \\ P= \sqrt{a^2+2ab+b^2}= \sqrt{c^2+4S}+c= \sqrt{12^2+4\times18}+12=6 \sqrt{6} +12\\ r= \frac{2S}{P}= \frac{2\cdot 18}{6 \sqrt{6} +12}= \frac{2\cdot 18(6\sqrt{6}-12)}{(6\sqrt{6})^2-12^2} =3\sqrt{6}-6\\ a= \frac{c+2r+ \sqrt{c^2-4cr-4r^2} }{2}= \frac{12+2(3\sqrt{6}-6)- \sqrt{12^2-4\cdot 12(3\sqrt{6}-6)-4(3\sqrt{6}-6)^2} }{2}=3\sqrt{6}+3\sqrt{2}](https://tex.z-dn.net/?f=P%3Da%2Bb%2Bc%5C%5C+P%3D+%5Csqrt%7B%28a%2Bb%29%5E2%7D%2Bc+%5C%5C+P%3D+%5Csqrt%7Ba%5E2%2B2ab%2Bb%5E2%7D%3D+%5Csqrt%7Bc%5E2%2B4S%7D%2Bc%3D+%5Csqrt%7B12%5E2%2B4%5Ctimes18%7D%2B12%3D6+%5Csqrt%7B6%7D+++++%2B12%5C%5C+r%3D+%5Cfrac%7B2S%7D%7BP%7D%3D+%5Cfrac%7B2%5Ccdot+18%7D%7B6+%5Csqrt%7B6%7D+%2B12%7D%3D+%5Cfrac%7B2%5Ccdot+18%286%5Csqrt%7B6%7D-12%29%7D%7B%286%5Csqrt%7B6%7D%29%5E2-12%5E2%7D+%3D3%5Csqrt%7B6%7D-6%5C%5C+a%3D+%5Cfrac%7Bc%2B2r%2B+%5Csqrt%7Bc%5E2-4cr-4r%5E2%7D+%7D%7B2%7D%3D+%5Cfrac%7B12%2B2%283%5Csqrt%7B6%7D-6%29-+%5Csqrt%7B12%5E2-4%5Ccdot+12%283%5Csqrt%7B6%7D-6%29-4%283%5Csqrt%7B6%7D-6%29%5E2%7D+%7D%7B2%7D%3D3%5Csqrt%7B6%7D%2B3%5Csqrt%7B2%7D)
![b= \frac{12+2(3\sqrt{6}-6)- \sqrt{12^2-4\cdot 12(3\sqrt{6}-6)-4(3\sqrt{6}-6)^2} }{2}=3\sqrt{6}-3\sqrt{2} b= \frac{12+2(3\sqrt{6}-6)- \sqrt{12^2-4\cdot 12(3\sqrt{6}-6)-4(3\sqrt{6}-6)^2} }{2}=3\sqrt{6}-3\sqrt{2}](https://tex.z-dn.net/?f=b%3D+%5Cfrac%7B12%2B2%283%5Csqrt%7B6%7D-6%29-+%5Csqrt%7B12%5E2-4%5Ccdot+12%283%5Csqrt%7B6%7D-6%29-4%283%5Csqrt%7B6%7D-6%29%5E2%7D+%7D%7B2%7D%3D3%5Csqrt%7B6%7D-3%5Csqrt%7B2%7D+)
![tg \beta= \frac{3\sqrt{6}-3\sqrt{2}}{3\sqrt{6}+3\sqrt{2}} =\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} =2-\sqrt{3} \\ \beta =arctg(2-\sqrt{3})=15а tg \beta= \frac{3\sqrt{6}-3\sqrt{2}}{3\sqrt{6}+3\sqrt{2}} =\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} =2-\sqrt{3} \\ \beta =arctg(2-\sqrt{3})=15а](https://tex.z-dn.net/?f=tg+%5Cbeta%3D++%5Cfrac%7B3%5Csqrt%7B6%7D-3%5Csqrt%7B2%7D%7D%7B3%5Csqrt%7B6%7D%2B3%5Csqrt%7B2%7D%7D++%3D%5Cfrac%7B%5Csqrt%7B6%7D-%5Csqrt%7B2%7D%7D%7B%5Csqrt%7B6%7D%2B%5Csqrt%7B2%7D%7D+%3D2-%5Csqrt%7B3%7D+%5C%5C++%5Cbeta+%3Darctg%282-%5Csqrt%7B3%7D%29%3D15%D0%B0)
![\alpha =90а-15а=75а \alpha =90а-15а=75а](https://tex.z-dn.net/?f=+%5Calpha+%3D90%D0%B0-15%D0%B0%3D75%D0%B0)
Окончательный ответ:
где
Окончательный ответ:
Приложения:
![](https://st.uroker.com/files/f55/f55c7c0ce1b899050801312c348b0c56.png)
Ответ дал:
1
решение во вложении, в файле
Приложения:
![](https://st.uroker.com/files/e0f/e0fd983a812f2ef30fc9cf134fe300d9.jpg)
![](https://st.uroker.com/files/fdd/fdde08c56afb5df70da3c23237983d87.jpg)
Вас заинтересует
11 месяцев назад
11 месяцев назад
1 год назад
6 лет назад
6 лет назад
7 лет назад