Obwód trapezu równoramiennego jest równy 24 cm, a cosinus jego kąta ostrego jest równy 3/5. Jakie powinny być długości boków tego trapezu, aby jego pole było równe 28cm2? Dzięki :)

L = 24 →L = a+b+2c→ a+b+2c = 24 cosα = 3/5 → cosα =(a-b)/2c→ (a-b)/2c = 3/5 P = 28 → P = 1/2(a+b)h → 1/2(a+b)h = 28 cos²α +sin²α = 1 sin²α = 1- 9/25 sin²α = 16/25 sinα = 4/5 z tw sinusów: h/sinα = c/sin90 h/(4/5) = c h = (4/5)c a+b+2c = 24 (a-b)/2c = 3/5 (4/10)c*(a+b) = 28 a+b+2c = 24 5(a+b) = 6c c(a+b) = 70 a+b = 24 - 2c 5(a+b) = 6c c(24 - 2c) = 70→ c² - 12c+ 35 = 0 Δ = 144 - 140 = 4 √Δ = 2 c = (12 - 2)/2 = 5 ∨c = (12 - 2)/2 = 7 c= 5 a+b = 14 5(a- b) = 30 c = 5 a = 14 - b 5(14 - 2b) = 30 c = 5 a = 14 - b 70 - 10b = 30 c = 5 a = 14-b -10b = -40 a = 10 b = 4 c = 5 lub: c= 7 a+b= 10 5(a- b) = 42 c= 7 a = 10 - 5 5(10 - 2b) = 42 c= 7 a = 10 - 5 50 - 10b = 42 c= 7 a = 10 - b b = 4/5 a = 46/5 b = 4/5 c = 7