pomocy !! KRYSTIAN: Wyznacz długosci boków i miary kątów trójkąta prostokątnego ABC (kąt przy wierzchołku C jesy prosty) wiedząc że |CAB| = 60stopni i |BC| = 8

Janek191: α = 60^o γ = 90^o więc β = 90^o − 60^o = 30^o a = I BC I = 8 b = I AC I c = ( AB I Mamy a/ b = tg α = tg 60^o = √3 a = b *√3 b = a/ √3 = 8/√3 = ( 8 √3) / 3 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− a / c = sin α = sin 60 ^o = √3/2 2a = √3 c c = (2a)/ √3 = 16/√3 = ( 16 √3 )/3 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

Janek191: α = 60^o γ = 90^o więc β = 90^o − 60^o = 30^o a = I BC I = 8 b = I AC I c = ( AB I Mamy a/ b = tg α = tg 60^o = √3 a = b *√3 b = a/ √3 = 8/√3 = ( 8 √3) / 3 −−−−−−−−−−−−−−−−−−−−−−−−−− a / c = sin α = sin 60 ^o = √3/2 2a = √3 c c = (2a)/ √3 = 16/√3 = ( 16 √3 )/3 −−−−−−−−−−−−−−−−−−−−−−−−−−−−