Given that a, b, c are the sides of a ∆ABC which is right angled at C, then the minimum value of ca+cb2 is
0
4
6
8
a=csinθ,b=ccosθ⇒ ca+cb2=1sinθ+1cosθ2=4(1+sin2θ)sin22θ=41sin22θ+1sin2θ, where 0<θ<π2⇒ ca+cbmin2=8, when 2θ=90∘.