Question

R²= c²+ (R/cosβ - b)²

proszę o pomoc w takim przekształceniu tego równania aby wyliczyć R . będzie to funkcja kwadratowa

Answer 1

R²= c²+ (R/cosβ - b)²
R²*cos²β = c²*cos²β + R² - 2Rb*cosβ + b²*cos²β
R²(cos²β - 1) + 2Rb*cosβ - (a² + b²)cos²β = 0
R²sin²β + 2Rb*cosβ - (c² + b²)cos²β = 0

Δ = 4b²cos²β + 4(c² + b²)cos²β*sin²β = 4cos²β(b² + (c² + b²)sin²β)
Δ ≥ 0

R = (- 2b*cosβ ± 2cosβ√(b² + (c² + b²)sin²β))/(2sin²β) = cosβ(- b ± √(b² + (c² + b²)sin²β))/(sin²β)

jak masz pytania to pisz na pw

Answer 2

R("2")=c("2")+(R/cosβ-b)("2")

R("2")x(razy)cos("2")β=c²*cos("2")β+R("2")-2Rb*cosβ+

b("2")x(razy)cos("2")β

R("2")(cos("2")β-1)+2Rbx(razy)cosβ-(a("2")+b("2"))cos("2")β=0

R("2")sin("2")β+2Rbx(razy)cosβ-(c("2")+b("2"))cos("2")β=0

/_\=4b("2")cos("2")β+4(c("2")+b("2"))cos("2")βx(razy)sin("2")β=

4cos("2")β(b("2")+(c("2")+b("2"))sin("2")β)

/_\ >_ 0

R=(-2bx(razy)cosβ±2cosβ√(b("2")+

(c("2")+b("2"))sin("2")β))/(2sin("2")β)=cosβ(-b±√(b("2")+

(c("2")+ b("2"))sin("2")β))/(sin("2")β)