Question

If $a,b,c$ be the sides of $△ABC$ and equations $a{x}^2+bx+c=0$ and $5x^2+12x+13=0$ have a common root, then $\angle C$ is

• A. ${45}^{\circ }$
• B. ${120}^{\circ }$
• C. ${90}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C ${90}^{\circ }$

Given: For $5x^2+12x+13=0$,
$\Delta ={12}^2-4\left(5\right)\left(13\right)=-116=-ve$

This means that the equation $5x^2+12x+13=0$ has complex roots ,i.e., both the roots will be common to $a{x}^2+bx+c=0$.

$\frac{a}{5}=\frac{b}{12}=\frac{c}{13}=k\left(Constant\right)$

$\Rightarrow a=5k,b=12k,c=13k$
Also, $(13k{)}^2=(5k{)}^2+(12k{)}^2$
$\Rightarrow c^2=a^2+b^2$

$△ABC$ is a right-angled triangle, right-angled at $C$.
$\Rightarrow \angle C={90}^{\circ }$