См. рисунок.
1) Пусть ABS - данный конус. SO⊥AB, SO - высота конуса, SO=H.
∠SBO=α.
S(полной поверхности)-?
2) S(п.п.)=Sоснования+Sбок.
So=πR²
Sб=πRl
Из треугольника SOB(∠SOB=90°)
OB=tg(α)*H=R
SB=H\sin(α)=l
[tex]Sp.p=S_o+S_b=\pi R^2+\pi Rl=\\\\ =\pi(tg\alpha *H)^2+\pi*tg\alpha*H*\frac{H}{sin\alpha }=\\\\=\pi tg^2(\alpha)H^2+\frac{\pi tg(\alpha)H^2}{sin\alpha }=\frac{\pi tg^2(\alpha )H^2 sin(\alpha )+\pi tg(\alpha )H^2}{sin\alpha }=\\\\={\frac{\pi tg(\alpha )H^2(tg(\alpha)sin(\alpha )+1)}{sin\alpha }=\frac{\pi*\frac{sin(\alpha)}{cos(\alpha )}H^2(\frac{sin(\alpha)}{cos(\alpha)}*sin(\alpha)+1 )}{sin(\alpha )}}=\\\\=\frac{\pi H^2\frac{sin(\alpha)}{cos(\alpha)}*\frac{sin^2(\alpha)+cos(\alpha)}{cos(\alpha)}}{sin(\alpha)}=\frac{\frac{\pi H^2(sin^2(\alpha)+cos(\alpha))}{cos^2(\alpha)}}{sin(\alpha)}=\frac{\pi H^2(sin^2(\alpha)+cos(\alpha))}{cos^2(\alpha)}=\\\\=\boxed{\frac{\pi H^2*sin^2(\alpha)+\pi H^2*cos(\alpha)}{cos^2(\alpha)} }[/tex]
