Sides of a triangle are 2 cm,sqrt{6} cm,and | vedclass

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(B) Let the sides be $a = 2$,$b = \sqrt{6}$,and $c = \sqrt{3} + 1$.We know that the smallest angle is opposite to the smallest side.Comparing the valu

Vedclass · Accepted Answer

$45$

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(B) Let the sides be $a = 2$,$b = \sqrt{6}$,and $c = \sqrt{3} + 1$.We know that the smallest angle is opposite to the smallest side.Comparing the values: $a = 2 \approx 2$,$b = \sqrt{6} \approx 2.45$,and $c = \sqrt{3} + 1 \approx 1.732 + 1 = 2.732$.The smallest side is $a = 2$.Using the Law of Cosines for angle $A$: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.$\cos A = \frac{(\sqrt{6})^2 + (\sqrt{3} + 1)^2 - 2^2}{2(\sqrt{6})(\sqrt{3} + 1)} = \frac{6 + (3 + 1 + 2\sqrt{3}) - 4}{2\sqrt{6}(\sqrt{3} + 1)} = \frac{6 + 2\sqrt{3}}{2\sqrt{6}(\sqrt{3} + 1)}$.$\cos A = \frac{2(3 + \sqrt{3})}{2\sqrt{6}(\sqrt{3} + 1)} = \frac{\sqrt{3}(\sqrt{3} + 1)}{\sqrt{6}(\sqrt{3} + 1)} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$.Since $\cos A = \frac{1}{\sqrt{2}}$,we have $A = 45^o$.

Sides of a triangle are $2 \ cm$,$\sqrt{6} \ cm$,and $(\sqrt{3} + 1) \ cm$. The smallest angle of the triangle is .....$^o$

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