In triangle ABC,if A=frac{pi}{3} and B=frac{p | vedclass

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Question

(A) Given $A=\frac{\pi}{3}$ and $B=\frac{\pi}{4}$. In $	riangle ABC$,$\angle C = \pi - (A+B) = \pi - (\frac{\pi}{3} + \frac{\pi}{4}) = \pi - \frac{7\

Vedclass · Accepted Answer

$2-\sqrt{3}$

Vedclass · Answer

(A) Given $A=\frac{\pi}{3}$ and $B=\frac{\pi}{4}$. In $	riangle ABC$,$\angle C = \pi - (A+B) = \pi - (\frac{\pi}{3} + \frac{\pi}{4}) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12}$.Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.Thus,$a = k \sin(\frac{\pi}{3}) = k \frac{\sqrt{3}}{2}$,$b = k \sin(\frac{\pi}{4}) = k \frac{1}{\sqrt{2}}$,and $c = k \sin(\frac{5\pi}{12}) = k \sin(\frac{\pi}{4} + \frac{\pi}{6}) = k (\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}) = k \frac{\sqrt{3}+1}{2\sqrt{2}}$.Now,$\frac{a^2-b^2}{c^2} = \frac{k^2(\frac{3}{4} - \frac{1}{2})}{k^2(\frac{(\sqrt{3}+1)^2}{8})} = \frac{1/4}{(\frac{3+1+2\sqrt{3}}{8})} = \frac{1/4}{(\frac{4+2\sqrt{3}}{8})} = \frac{1/4}{(\frac{2+\sqrt{3}}{4})} = \frac{1}{2+\sqrt{3}}$.Rationalizing the denominator: $\frac{1}{2+\sqrt{3}} 	imes \frac{2-\sqrt{3}}{2-\sqrt{3}} = 2-\sqrt{3}$.Thus,the correct option is $A$.

In triangle $ABC$,if $A=\frac{\pi}{3}$ and $B=\frac{\pi}{4}$,then $\frac{a^2-b^2}{c^2}=$

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