ધારો કે mathrm{ABC} એ એક સમદ્વિબાજુ ત્રિકોણ છ | vedclass

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Question

$\frac{\mathrm{c}}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}}$ [By sine rule]

$2 c=8 \Rightarrow c=4$

$ \mathrm{AB}=|(\mathrm{b}+1)|=4

Vedclass · Accepted Answer

$36$

Vedclass · Answer

$\frac{\mathrm{c}}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}}$ [By sine rule]

$2 c=8 \Rightarrow c=4$

$ \mathrm{AB}=|(\mathrm{b}+1)|=4 $

$ \mathrm{~b}=3, \mathrm{~m}_{\mathrm{AB}}=0 $

$ \mathrm{~m}_{\mathrm{BC}}=\frac{-1}{\sqrt{3}} $

$ B C:-y=\frac{-1}{\sqrt{3}}(x-3) $

$ \sqrt{3} \mathrm{y}+\mathrm{x}=3 $

$ 	ext { Point of intersection : } y=x+3, \sqrt{3} y+x=3 $

$ (\sqrt{3}+1) \mathrm{y}=6 $

$ \mathrm{y}=\frac{6}{\sqrt{3}+1} $

$ x=\frac{6}{\sqrt{3}+1}-3 $

$ =\frac{6-3 \sqrt{3}-3}{\sqrt{3}+1} $

$ =3 \frac{(1-\sqrt{3})}{(1+\sqrt{3})}=\frac{-6}{(1+\sqrt{3})^2} $

$ \frac{\beta^4}{\alpha^2}=36 $

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