In a triangle ABC,sin A and sin B satisfy the | vedclass

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(D) Given the quadratic equation $c^2 x^2 - c(a+b)x + ab = 0$. Factoring the equation: $c^2 x^2 - cax - cbx + ab = 0$ $cx(cx - a) - b(cx - a) = 0$ $(c

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$\sin A + \cos A = \frac{a+b}{c}$

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(D) Given the quadratic equation $c^2 x^2 - c(a+b)x + ab = 0$. Factoring the equation: $c^2 x^2 - cax - cbx + ab = 0$ $cx(cx - a) - b(cx - a) = 0$ $(cx - a)(cx - b) = 0$ Thus,the roots are $x = \frac{a}{c}$ and $x = \frac{b}{c}$. Since $\sin A$ and $\sin B$ are the roots,we have $\sin A = \frac{a}{c}$ and $\sin B = \frac{b}{c}$. From the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Substituting $\sin A = \frac{a}{c}$ and $\sin B = \frac{b}{c}$,we get $c = \frac{a}{\sin A}$ and $c = \frac{b}{\sin B}$. This implies $\frac{a}{\sin A} = c$ and $\frac{b}{\sin B} = c$. Comparing with the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$,we get $\frac{c}{\sin C} = c$,which means $\sin C = 1$. Therefore,$\angle C = 90^\circ$,so the triangle is right-angled at $C$. In a right-angled triangle at $C$,$\sin A = \frac{a}{c}$ and $\cos A = \sin B = \frac{b}{c}$. Thus,$\sin A + \cos A = \frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}$.

In a $\triangle ABC$,$\sin A$ and $\sin B$ satisfy the equation $c^2 x^2 - c(a+b)x + ab = 0$. Then:

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