Question

Given that $a,b,c$ are the sides of a triangle $ABC$ which is right angled at $C$, then the minimum value of ${(\frac{c}{a}+\frac{c}{b})}^2$ is

• A. $0$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$


$a=c\sin \theta ,b=c\cos \theta$
$\Rightarrow {(\frac{c}{a}+\frac{c}{b})}^2={(\frac{1}{\sin \theta }+\frac{1}{\cos \theta })}^2$
$=\frac{4(1+\sin 2\theta )}{{\sin }^{2}2\theta }$
$=4(\frac{1}{{\sin }^{2}2\theta }+\frac{1}{\sin 2\theta })$, where $0<\theta <\frac{\pi }{2}$
$\Rightarrow {(\frac{c}{a}+\frac{c}{b})}_{min}^2=8$, when $2\theta ={90}^{\circ }$