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 $C$ is:

• A. $60$
• B. $90$
• C. $120$
• D. $45$

Answer 1

The correct option is B $90$

If $a{x}^2+bx+c$ and $5x^2+12x+13=0$ have a common root, then we have

$\frac{a}{5}=\frac{b}{12}=\frac{c}{13}=k$

$\Rightarrow a=5K.b=12K,c=13K$

and $(13K{)}^2=(12K{)}^2+(25K{)}^2$

Since $a,bandc$ are sides of $\Delta ABC$, and follows Pythagoras theorem.

$\therefore \Delta ABC$ is right angled triangle

Now using angle $C$ by $\cos C$

$\cos C=\frac{a^2+b^2-c^2}{2ab}=0$

$\Rightarrow C={90}^{\circ }$