По формуле двойного угла
sin(2a)=2*sina*cosa
![8+\frac{8\sin 36^0}{\sin 72^0}=8+\frac{8\sin 36^0}{2\sin 36^0\cos 36^0}](https://tex.z-dn.net/?f=8%2B%5Cfrac%7B8%5Csin%2036%5E0%7D%7B%5Csin%2072%5E0%7D%3D8%2B%5Cfrac%7B8%5Csin%2036%5E0%7D%7B2%5Csin%2036%5E0%5Ccos%2036%5E0%7D)
![8+\frac{8\sin 36^0}{2\sin 36^0\cos 36^0}=8+\frac{4}{\cos 36^0}](https://tex.z-dn.net/?f=8%2B%5Cfrac%7B8%5Csin%2036%5E0%7D%7B2%5Csin%2036%5E0%5Ccos%2036%5E0%7D%3D8%2B%5Cfrac%7B4%7D%7B%5Ccos%2036%5E0%7D)
Известно, что ![\cos36^0=\frac{1+\sqrt{5}}{4}](https://tex.z-dn.net/?f=%5Ccos36%5E0%3D%5Cfrac%7B1%2B%5Csqrt%7B5%7D%7D%7B4%7D)
![8+\frac{4}{\cos 36^0}=8+\frac{4}{\frac{1+\sqrt{5}}{4}}](https://tex.z-dn.net/?f=8%2B%5Cfrac%7B4%7D%7B%5Ccos%2036%5E0%7D%3D8%2B%5Cfrac%7B4%7D%7B%5Cfrac%7B1%2B%5Csqrt%7B5%7D%7D%7B4%7D%7D)
![8+\frac{4}{\frac{1+\sqrt{5}}{4}}=8+\frac{16}{1+\sqrt{5}}](https://tex.z-dn.net/?f=8%2B%5Cfrac%7B4%7D%7B%5Cfrac%7B1%2B%5Csqrt%7B5%7D%7D%7B4%7D%7D%3D8%2B%5Cfrac%7B16%7D%7B1%2B%5Csqrt%7B5%7D%7D)
![8+\frac{16}{1+\sqrt{5}}=8+\frac{16*(\sqrt{5}-1)}{(1+\sqrt{5})*(\sqrt{5}-1)}](https://tex.z-dn.net/?f=8%2B%5Cfrac%7B16%7D%7B1%2B%5Csqrt%7B5%7D%7D%3D8%2B%5Cfrac%7B16%2A%28%5Csqrt%7B5%7D-1%29%7D%7B%281%2B%5Csqrt%7B5%7D%29%2A%28%5Csqrt%7B5%7D-1%29%7D)
![8+\frac{16*(\sqrt{5}-1)}{(1+\sqrt{5})*(\sqrt{5}-1)}=8+\frac{16*(\sqrt{5}-1)}{(\sqrt{5})^2-1^2}](https://tex.z-dn.net/?f=8%2B%5Cfrac%7B16%2A%28%5Csqrt%7B5%7D-1%29%7D%7B%281%2B%5Csqrt%7B5%7D%29%2A%28%5Csqrt%7B5%7D-1%29%7D%3D8%2B%5Cfrac%7B16%2A%28%5Csqrt%7B5%7D-1%29%7D%7B%28%5Csqrt%7B5%7D%29%5E2-1%5E2%7D)
![8+\frac{16*(\sqrt{5}-1)}{(\sqrt{5})^2-1^2}=8+\frac{16*(\sqrt{5}-1)}{4}](https://tex.z-dn.net/?f=8%2B%5Cfrac%7B16%2A%28%5Csqrt%7B5%7D-1%29%7D%7B%28%5Csqrt%7B5%7D%29%5E2-1%5E2%7D%3D8%2B%5Cfrac%7B16%2A%28%5Csqrt%7B5%7D-1%29%7D%7B4%7D)
![8+\frac{16*(\sqrt{5}-1)}{4}=8+4(\sqrt{5}-1)](https://tex.z-dn.net/?f=8%2B%5Cfrac%7B16%2A%28%5Csqrt%7B5%7D-1%29%7D%7B4%7D%3D8%2B4%28%5Csqrt%7B5%7D-1%29)
![8+4(\sqrt{5}-1)=4+4\sqrt{5}](https://tex.z-dn.net/?f=8%2B4%28%5Csqrt%7B5%7D-1%29%3D4%2B4%5Csqrt%7B5%7D)
![4+4\sqrt{5}=4*(1+\sqrt{5})](https://tex.z-dn.net/?f=4%2B4%5Csqrt%7B5%7D%3D4%2A%281%2B%5Csqrt%7B5%7D%29)
Во втором примере воспользуемся формулой
![\tan\alpha=\frac{\sin\alpha}{\cos\alpha}](https://tex.z-dn.net/?f=%5Ctan%5Calpha%3D%5Cfrac%7B%5Csin%5Calpha%7D%7B%5Ccos%5Calpha%7D)
![8*\cos 36^0+\frac{\tan 72^0}{8*\sin 36^0}=8*\frac{\sqrt{5}+1}{4}+\frac{\frac{\sin 72^0}{\cos 72^0}}{8*\sin 36^0}](https://tex.z-dn.net/?f=8%2A%5Ccos%2036%5E0%2B%5Cfrac%7B%5Ctan%2072%5E0%7D%7B8%2A%5Csin%2036%5E0%7D%3D8%2A%5Cfrac%7B%5Csqrt%7B5%7D%2B1%7D%7B4%7D%2B%5Cfrac%7B%5Cfrac%7B%5Csin%2072%5E0%7D%7B%5Ccos%2072%5E0%7D%7D%7B8%2A%5Csin%2036%5E0%7D)
![8*\frac{\sqrt{5}+1}{4}+\frac{\frac{\sin 72^0}{\cos 72^0}}{8*\sin 36^0}=2*(\sqrt{5}+1)+\frac{\sin 72^0}{8\sin 36^0\cos 72^0}](https://tex.z-dn.net/?f=8%2A%5Cfrac%7B%5Csqrt%7B5%7D%2B1%7D%7B4%7D%2B%5Cfrac%7B%5Cfrac%7B%5Csin%2072%5E0%7D%7B%5Ccos%2072%5E0%7D%7D%7B8%2A%5Csin%2036%5E0%7D%3D2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%5Csin%2072%5E0%7D%7B8%5Csin%2036%5E0%5Ccos%2072%5E0%7D)
![2*(\sqrt{5}+1)+\frac{\sin 72^0}{8\sin 36^0\cos 72^0}=2*(\sqrt{5}+1)+\frac{2\sin 36^0\cos 36^0}{8\sin 36^0\cos 72^0}](https://tex.z-dn.net/?f=2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%5Csin%2072%5E0%7D%7B8%5Csin%2036%5E0%5Ccos%2072%5E0%7D%3D2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B2%5Csin%2036%5E0%5Ccos%2036%5E0%7D%7B8%5Csin%2036%5E0%5Ccos%2072%5E0%7D)
![2*(\sqrt{5}+1)+\frac{2\sin 36^0\cos 36^0}{8\sin 36^0\cos 72^0}=2*(\sqrt{5}+1)+\frac{\cos 36^0}{4\cos 72^0}\quad(2)](https://tex.z-dn.net/?f=2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B2%5Csin%2036%5E0%5Ccos%2036%5E0%7D%7B8%5Csin%2036%5E0%5Ccos%2072%5E0%7D%3D2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%5Ccos%2036%5E0%7D%7B4%5Ccos%2072%5E0%7D%5Cquad%282%29)
Вычислим отдельно ![\cos 72^0=\cos(2*36^0)](https://tex.z-dn.net/?f=%5Ccos%2072%5E0%3D%5Ccos%282%2A36%5E0%29)
По формуле двойного угла для косинуса
![\cos2\alpha=2\cos^2\alpha-1](https://tex.z-dn.net/?f=%5Ccos2%5Calpha%3D2%5Ccos%5E2%5Calpha-1)
![\cos(2*36^0)=2\cos^2 36^0-1](https://tex.z-dn.net/?f=%5Ccos%282%2A36%5E0%29%3D2%5Ccos%5E2%2036%5E0-1)
![2\cos^2 36^0-1=2*\left(\frac{\sqrt{5}+1}{4}\right)^2-1](https://tex.z-dn.net/?f=2%5Ccos%5E2%2036%5E0-1%3D2%2A%5Cleft%28%5Cfrac%7B%5Csqrt%7B5%7D%2B1%7D%7B4%7D%5Cright%29%5E2-1)
![2*\left(\frac{\sqrt{5}+1}{4}\right)^2-1=\frac{(\sqrt{5}+1)^2}{8}-1](https://tex.z-dn.net/?f=2%2A%5Cleft%28%5Cfrac%7B%5Csqrt%7B5%7D%2B1%7D%7B4%7D%5Cright%29%5E2-1%3D%5Cfrac%7B%28%5Csqrt%7B5%7D%2B1%29%5E2%7D%7B8%7D-1)
![\frac{(\sqrt{5}+1)^2}{8}-1=\frac{6+2\sqrt{5}}{8}-1](https://tex.z-dn.net/?f=%5Cfrac%7B%28%5Csqrt%7B5%7D%2B1%29%5E2%7D%7B8%7D-1%3D%5Cfrac%7B6%2B2%5Csqrt%7B5%7D%7D%7B8%7D-1)
![\frac{6+2\sqrt{5}}{8}-1=\frac{3+\sqrt{5}}{4}-1](https://tex.z-dn.net/?f=%5Cfrac%7B6%2B2%5Csqrt%7B5%7D%7D%7B8%7D-1%3D%5Cfrac%7B3%2B%5Csqrt%7B5%7D%7D%7B4%7D-1)
![\frac{3+\sqrt{5}}{4}-1=\frac{3+\sqrt{5}-4}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2B%5Csqrt%7B5%7D%7D%7B4%7D-1%3D%5Cfrac%7B3%2B%5Csqrt%7B5%7D-4%7D%7B4%7D)
![\frac{3+\sqrt{5}-4}{4}=\frac{\sqrt{5}-1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2B%5Csqrt%7B5%7D-4%7D%7B4%7D%3D%5Cfrac%7B%5Csqrt%7B5%7D-1%7D%7B4%7D)
Значит
![\cos 72^0=\frac{\sqrt{5}-1}{4}](https://tex.z-dn.net/?f=%5Ccos%2072%5E0%3D%5Cfrac%7B%5Csqrt%7B5%7D-1%7D%7B4%7D)
Вернемся к (2)
![2*(\sqrt{5}+1)+\frac{\cos 36^0}{4\cos 72^0}=2*(\sqrt{5}+1)+\frac{\cos 36^0}{\sqrt{5}-1}](https://tex.z-dn.net/?f=2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%5Ccos%2036%5E0%7D%7B4%5Ccos%2072%5E0%7D%3D2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%5Ccos%2036%5E0%7D%7B%5Csqrt%7B5%7D-1%7D)
![2*(\sqrt{5}+1)+\frac{\cos 36^0}{\sqrt{5}-1}=2*(\sqrt{5}+1)+\frac{\sqrt{5}+1}{4(\sqrt{5}-1)}](https://tex.z-dn.net/?f=2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%5Ccos%2036%5E0%7D%7B%5Csqrt%7B5%7D-1%7D%3D2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%5Csqrt%7B5%7D%2B1%7D%7B4%28%5Csqrt%7B5%7D-1%29%7D)
![2*(\sqrt{5}+1)+\frac{\sqrt{5}+1}{4(\sqrt{5}-1)}=2*(\sqrt{5}+1)+\frac{(\sqrt{5}+1)^2}{4(\sqrt{5}-1)*(\sqrt{5}+1)}](https://tex.z-dn.net/?f=2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%5Csqrt%7B5%7D%2B1%7D%7B4%28%5Csqrt%7B5%7D-1%29%7D%3D2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%28%5Csqrt%7B5%7D%2B1%29%5E2%7D%7B4%28%5Csqrt%7B5%7D-1%29%2A%28%5Csqrt%7B5%7D%2B1%29%7D)
![2*(\sqrt{5}+1)+\frac{(\sqrt{5}+1)^2}{4(\sqrt{5}-1)*(\sqrt{5}+1)}=2*(\sqrt{5}+1)+\frac{(\sqrt{5}+1)^2}{4*4}](https://tex.z-dn.net/?f=2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%28%5Csqrt%7B5%7D%2B1%29%5E2%7D%7B4%28%5Csqrt%7B5%7D-1%29%2A%28%5Csqrt%7B5%7D%2B1%29%7D%3D2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%28%5Csqrt%7B5%7D%2B1%29%5E2%7D%7B4%2A4%7D)
![2*(\sqrt{5}+1)+\frac{(\sqrt{5}+1)^2}{4*4}=2*(\sqrt{5}+1)+\frac{(6+2\sqrt{5})}{4*4}](https://tex.z-dn.net/?f=2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%28%5Csqrt%7B5%7D%2B1%29%5E2%7D%7B4%2A4%7D%3D2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%286%2B2%5Csqrt%7B5%7D%29%7D%7B4%2A4%7D)
![2*(\sqrt{5}+1)+\frac{(6+2\sqrt{5})}{4*4}=2*(\sqrt{5}+1)+\frac{(3+\sqrt{5})}{8}](https://tex.z-dn.net/?f=2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%286%2B2%5Csqrt%7B5%7D%29%7D%7B4%2A4%7D%3D2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%283%2B%5Csqrt%7B5%7D%29%7D%7B8%7D)
![2*(\sqrt{5}+1)+\frac{(3+\sqrt{5})}{8}=\frac{(16+3+16\sqrt{5}+\sqrt{5})}{8}](https://tex.z-dn.net/?f=2%2A%28%5Csqrt%7B5%7D%2B1%29%2B%5Cfrac%7B%283%2B%5Csqrt%7B5%7D%29%7D%7B8%7D%3D%5Cfrac%7B%2816%2B3%2B16%5Csqrt%7B5%7D%2B%5Csqrt%7B5%7D%29%7D%7B8%7D)
![\frac{(16+3+16\sqrt{5}+\sqrt{5})}{8}=\frac{(19+17\sqrt{5})}{8}](https://tex.z-dn.net/?f=%5Cfrac%7B%2816%2B3%2B16%5Csqrt%7B5%7D%2B%5Csqrt%7B5%7D%29%7D%7B8%7D%3D%5Cfrac%7B%2819%2B17%5Csqrt%7B5%7D%29%7D%7B8%7D)
S1=(3+7)/2*3=15
S2=(3+7)/2*3=15
Sобщ.=15+15=30
Ответ:30
<span>По теореме Пифагора:</span>
тр. ОБС (С- точка касания) - прямоугольный, т.к. касательная всегда перпендикулярна радиусу.
СО=5 (радиус)
ВО=13 (гипотенуза)
СВ в кв. = 169 - 25= 144
СВ=12
Доказываем, что тр.ОВС= тр. АСО по 2-м сторонам и углу между ними.
Следовательно, АС=СВ=12
АВ=24
<span>В сечении имеем шестиугольник.
Две стороны сечения призмы, проходящего через середины ребер AB, AD, B1C1, это отрезки длиной 2</span>√2<span>.
Боковые стороны равны </span>√(2²+3²) =√(4+9) = √13.
<span>Наклонная длина шестиугольника равна L = </span>√(6²+(2√2)²) = √(36+8) = √44 = 2√11.<span>
Ширина его по диагонали, параллельной основаниям, равна диагонали основания призмы В = 4</span>√2.
Сечение состоит из двух трапеций с равными основаниями.
S = Вср*L = ((2√2+4√2)/2)*2√11 = 3√2*2√11 = 6√22 кв.ед.