If the sides of a triangle are in the ratio 2 | vedclass

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(B) Let the sides of the triangle be $a = 2$,$b = \sqrt{6}$,and $c = \sqrt{3} + 1$.Since $c = \sqrt{3} + 1 \approx 1.732 + 1 = 2.732$,$b = \sqrt{6} \a

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) Let the sides of the triangle be $a = 2$,$b = \sqrt{6}$,and $c = \sqrt{3} + 1$.Since $c = \sqrt{3} + 1 \approx 1.732 + 1 = 2.732$,$b = \sqrt{6} \approx 2.449$,and $a = 2$,the largest side is $c$.Therefore,the largest angle is $C$,opposite to side $c$.Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.$\cos C = \frac{2^2 + (\sqrt{6})^2 - (\sqrt{3} + 1)^2}{2 	imes 2 	imes \sqrt{6}}$.$\cos C = \frac{4 + 6 - (3 + 1 + 2\sqrt{3})}{4\sqrt{6}} = \frac{10 - 4 - 2\sqrt{3}}{4\sqrt{6}} = \frac{6 - 2\sqrt{3}}{4\sqrt{6}}$.$\cos C = \frac{2\sqrt{3}(\sqrt{3} - 1)}{4\sqrt{6}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$.This value corresponds to $\cos(75^o)$.Thus,the largest angle is $75^o$.

If the sides of a triangle are in the ratio $2 : \sqrt{6} : (\sqrt{3} + 1)$,then the largest angle of the triangle will be.....$^o$

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