In triangle ABC, right-angled at A, the circu | vedclass

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(C) Given that $\angle A = 90^{\circ}$.Circumradius $R = \frac{a}{2 \sin A} = \frac{a}{2 \sin 90^{\circ}} = \frac{a}{2}$.Inradius $r = (s-a) 	an \fra

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$1, 3$

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(C) Given that $\angle A = 90^{\circ}$.Circumradius $R = \frac{a}{2 \sin A} = \frac{a}{2 \sin 90^{\circ}} = \frac{a}{2}$.Inradius $r = (s-a) 	an \frac{A}{2} = (s-a) 	an 45^{\circ} = s-a$.Exradius $r_1 = s 	an \frac{A}{2} = s 	an 45^{\circ} = s$.The ratio $R:r:r_1 = \frac{a}{2} : s-a : s = 2:5:\lambda$.From $\frac{a}{2} = 2$, we get $a = 4$.From $s-a = 5$, we get $s-4 = 5$, so $s = 9$.Thus, $\lambda = s = 9$.The equation becomes $x^2 - (9-5)x + (9-6) = 0$, which is $x^2 - 4x + 3 = 0$.Factoring gives $(x-1)(x-3) = 0$.Therefore, the roots are $1, 3$.

In $\triangle ABC$, right-angled at $A$, the circumradius, inradius, and radius of the excircle opposite to $A$ are respectively in the ratio $2:5:\lambda$. Then the roots of the equation $x^2-(\lambda-5)x+(\lambda-6)=0$ are:

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