Odpowiedź:
Tw. kosinusów
12² = 10² + a² - 2*10*a*cos 120°
144 = 100 + a² - 20 a*( - sin 30° )
144 = 100 + a² + 20 a*0,5
a² + 10 a - 44 = 0
-------------------------
Δ = 10² - 4*1*( -44) = 100 + 176 = 276 = 4*69
√Δ = 2[tex]\sqrt{69}[/tex]
a = [tex]\frac{- 10 + 2\sqrt{69} }{2} = \sqrt{69} - 5[/tex]
zatem
PΔ = 0,5*10*([tex]\sqrt{69}- 5) *sin 120[/tex]° = 5*([tex]\sqrt{69} - 5)*cos 30^o =[/tex]
= 5*([tex]\sqrt{69} - 5)*\frac{\sqrt{3} }{2} =[/tex] 2,5*([tex]\sqrt{69} - 5)*\sqrt{3}[/tex]
=========================================
Szczegółowe wyjaśnienie:
PΔ = 0,5 a*b*sin α
---------------------------------
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Verified answer
Odpowiedź:
Tw. kosinusów
12² = 10² + a² - 2*10*a*cos 120°
144 = 100 + a² - 20 a*( - sin 30° )
144 = 100 + a² + 20 a*0,5
a² + 10 a - 44 = 0
-------------------------
Δ = 10² - 4*1*( -44) = 100 + 176 = 276 = 4*69
√Δ = 2[tex]\sqrt{69}[/tex]
a = [tex]\frac{- 10 + 2\sqrt{69} }{2} = \sqrt{69} - 5[/tex]
zatem
PΔ = 0,5*10*([tex]\sqrt{69}- 5) *sin 120[/tex]° = 5*([tex]\sqrt{69} - 5)*cos 30^o =[/tex]
= 5*([tex]\sqrt{69} - 5)*\frac{\sqrt{3} }{2} =[/tex] 2,5*([tex]\sqrt{69} - 5)*\sqrt{3}[/tex]
=========================================
Szczegółowe wyjaśnienie:
PΔ = 0,5 a*b*sin α
---------------------------------