The sides of a triangle are 6+2 sqrt{3},4 sqr | vedclass

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Question

(A) Given the sides of the triangle are $a = 6+2 \sqrt{3}$,$b = 4 \sqrt{3}$,and $c = \sqrt{24} = 2 \sqrt{6}$.Comparing the values: $a \approx 6 + 3.46

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$\frac{1}{\sqrt{3}}$

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(A) Given the sides of the triangle are $a = 6+2 \sqrt{3}$,$b = 4 \sqrt{3}$,and $c = \sqrt{24} = 2 \sqrt{6}$.Comparing the values: $a \approx 6 + 3.464 = 9.464$,$b \approx 4 	imes 1.732 = 6.928$,and $c \approx 2 	imes 2.449 = 4.898$.Since $c$ is the smallest side,the smallest angle is $C$.Using the law of cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.$a^2 = (6+2 \sqrt{3})^2 = 36 + 12 + 24 \sqrt{3} = 48 + 24 \sqrt{3}$.$b^2 = (4 \sqrt{3})^2 = 48$.$c^2 = 24$.$\cos C = \frac{48 + 24 \sqrt{3} + 48 - 24}{2(6+2 \sqrt{3})(4 \sqrt{3})} = \frac{72 + 24 \sqrt{3}}{8 \sqrt{3}(3+\sqrt{3})} = \frac{24(3+\sqrt{3})}{8 \sqrt{3}(3+\sqrt{3})} = \frac{3}{\sqrt{3}} = \sqrt{3}$.Wait,checking the calculation again: $\cos C = \frac{72+24 \sqrt{3}}{16 \sqrt{3}(3+\sqrt{3})} = \frac{24(3+\sqrt{3})}{16 \sqrt{3}(3+\sqrt{3})} = \frac{24}{16 \sqrt{3}} = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}$.Thus,$\cos C = \frac{\sqrt{3}}{2}$,which means $C = 30^{\circ}$.The tangent of the smallest angle is $	an 30^{\circ} = \frac{1}{\sqrt{3}}$.

The sides of a triangle are $6+2 \sqrt{3}$,$4 \sqrt{3}$,and $\sqrt{24}$. The tangent of the smallest angle of the triangle is

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