In a triangle ABC, if a: b: c = 4: 5: 6, then | vedclass

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(A) Given $a: b: c = 4: 5: 6$. Let $a = 4x, b = 5x, c = 6x$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{4x+5x+6x}{2} = \frac{15x}{2}$.The circumra

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$16: 7$

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(A) Given $a: b: c = 4: 5: 6$. Let $a = 4x, b = 5x, c = 6x$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{4x+5x+6x}{2} = \frac{15x}{2}$.The circumradius $R = \frac{abc}{4\Delta}$ and the inradius $r = \frac{\Delta}{s}$, where $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.The ratio $\frac{R}{r} = \frac{abc}{4\Delta} 	imes \frac{s}{\Delta} = \frac{abcs}{4\Delta^2} = \frac{abcs}{4s(s-a)(s-b)(s-c)} = \frac{abc}{4(s-a)(s-b)(s-c)}$.Substituting the values:$(s-a) = \frac{15x}{2} - 4x = \frac{7x}{2}$, $(s-b) = \frac{15x}{2} - 5x = \frac{5x}{2}$, $(s-c) = \frac{15x}{2} - 6x = \frac{3x}{2}$.$\frac{R}{r} = \frac{(4x)(5x)(6x)}{4(\frac{7x}{2})(\frac{5x}{2})(\frac{3x}{2})} = \frac{120x^3}{4 	imes \frac{105x^3}{8}} = \frac{120x^3}{\frac{105x^3}{2}} = \frac{240}{105} = \frac{16}{7}$.Thus, the ratio $R: r = 16: 7$.

In a $\triangle ABC$, if $a: b: c = 4: 5: 6$, then the ratio of the radius of its circumcircle to that of its incircle is:

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