The equation 5x^2+12x + 13 = 0 and ax^2+bx + | vedclass

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Question

Roots of $5 \mathrm{x}^{2}+12 \mathrm{x}+13=0$ are Imaginary so

$\frac{a}{5}=\frac{b}{12}=\frac{c}{13}=k(L e t)$

So $\cos C=\frac{a^{2}+b^{2}-c^

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Vedclass · Answer

Roots of $5 \mathrm{x}^{2}+12 \mathrm{x}+13=0$ are Imaginary so

$\frac{a}{5}=\frac{b}{12}=\frac{c}{13}=k(L e t)$

So $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$

$\cos C=\frac{(5 k)^{2}+(12 k)^{2}-(13 k)^{2}}{2(5 k)(12 k)}=0$

$\cos C=0 \Rightarrow \angle C=\frac{\pi}{2}$

The equation $5x^2+12x + 13 = 0$ and $ax^2+bx + c = 0$ have a common root, where $a,b,c$ are the sides of $\Delta ABC$,then find $\angle C$ ? .....$^o$

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