Question

Answer each of the following questions in one word or one sentence or as per exact requirement of for question:

$Ina\Delta ABC,ifsinAandsinB\text{are the roots of the equation}{c}^2{x}^2-c(a+b)x+ab=0,thenfind\angle C.$

Answer 1

$c^2{x}^2-c(a+b)x+ab=0\text{sin A and sin B are the roots of the equation,}sinA+sinB=\frac{c(a+b)}{c^2}=\frac{(a+b)}{c}sinAsinB=\frac{ab}{c^2}sinA-sinB=\sqrt{{(sinA+sinB)}^2-4sinA.sinB}=\sqrt{{\left(\frac{a+b}{c}\right)}^2-4\frac{ab}{c^2}}=\sqrt{\frac{a^2+b^2+2ab-4ab}{c^2}}=\sqrt{\frac{a^2+b^2-2ab}{c^2}}=\sqrt{\frac{{(a-b)}^2}{c^2}}=\frac{left(a-b)}{c}sinA+sinB=\frac{(a+b)}{c}...\left(i\right)sinA-sinB=\frac{(a-b)}{c}...\left(ii\right)\text{Adding (i) and (ii), we get,}sinA=\frac{a}{c}andsinB=\frac{b}{c}sinA=\frac{Perpendicular}{hypotenuse}So,In\Delta ABC,\text{one of the side = a, another side = b,}\text{Third side = c}\therefore a^2+b^2=c^2\angle C={90}^{\circ }$