In a triangle ABC,A = 30^{circ} + C and R = ( | vedclass

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(C) Given $A = 30^{\circ} + C$ and $R = (\sqrt{3} + 1)r$. We know that $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$. Substituting $r

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$ABC$ is acute-angled

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(C) Given $A = 30^{\circ} + C$ and $R = (\sqrt{3} + 1)r$. We know that $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$. Substituting $r = \frac{R}{\sqrt{3} + 1}$,we get $\frac{1}{\sqrt{3} + 1} = 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$. Rationalizing the denominator,$\frac{\sqrt{3} - 1}{2} = 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$,so $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{\sqrt{3} - 1}{8}$. Using $A + B + C = 180^{\circ}$,we have $B = 180^{\circ} - (A + C) = 180^{\circ} - (30^{\circ} + 2C) = 150^{\circ} - 2C$. Substituting these values into the identity and solving for the angles,we find $\angle A = 75^{\circ}, \angle B = 60^{\circ}, \angle C = 45^{\circ}$. Since $\angle B = 60^{\circ}$ and $\angle A + \angle C = 120^{\circ}$,the triangle satisfies the given conditions. Thus,the triangle is not right-angled,equilateral,or acute-angled (since $75^{\circ} > 90^{\circ}$ is false,but $75^{\circ}$ is acute,$60^{\circ}$ is acute,$45^{\circ}$ is acute). Wait,$75^{\circ}, 60^{\circ}, 45^{\circ}$ are all acute,so $ABC$ is acute-angled.

In a $\triangle ABC$,$A = 30^{\circ} + C$ and $R = (\sqrt{3} + 1)r$,where $r$ is the inradius and $R$ is the circumradius. Then:

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