Объяснение:
BD = h, ∠DBC=α , AB = AC = x, BD - высота, S=?
1) BD - высота => BD⊥AC => ∠BDC = ∠BDA = 90°
2) В ΔBDC ∠BDC=90° => ∠BCD = 90° - α
3) В ΔBAC ∠BAC = 180° - ∠ABC - ∠BCA = 180° - 2∠BCA = 180° - 2(90° - α) = 180° - 180° + 2α = 2α
4) В ΔBDA ∠BDA=90° =>
5)
Объяснение:
BD = h, ∠DBC=α , AB = AC = x, BD - высота, S=?
1) BD - высота => BD⊥AC => ∠BDC = ∠BDA = 90°
2) В ΔBDC ∠BDC=90° => ∠BCD = 90° - α
3) В ΔBAC ∠BAC = 180° - ∠ABC - ∠BCA = 180° - 2∠BCA = 180° - 2(90° - α) = 180° - 180° + 2α = 2α
4) В ΔBDA ∠BDA=90° =>![sinBAD = \frac{BD}{AB} = sin(2\alpha ) = AB = \frac{BD}{sin(2\alpha )} = \frac{h}{sin(2\alpha )}](/tpl/images/0896/7042/432cd.png)
5)![S_{BAC} = \frac{1}{2} * BD*AC = \frac{1}{2} * BD*AB = \frac{1}{2} * h*\frac{h}{sin(2\alpha )} = \frac{h^{2} }{2*sin(2\alpha )} = \frac{h^{2} }{4*sin(\alpha)*cos(\alpha)}](/tpl/images/0896/7042/9d3d3.png)