- Home
- Standard 11
- Mathematics
ધારો કે $\mathrm{ABC}$ એ એક સમદ્વિબાજુ ત્રિકોણ છે, જેમાં $\mathrm{A}$ એ $(-1,0)$ આગળ છે, $\angle \mathrm{A}=\frac{2 \pi}{3}, \mathrm{AB}=\mathrm{AC}$ અને $\mathrm{B}$ એ ધન $x$-અક્ષ પર આવેલી છે. જો $\mathrm{BC}=4 \sqrt{3}$ અને રેખા $\mathrm{BC}$ એ, રેખા $y=x+3$ ને $(\alpha, \beta)$ આગળ છેદે તો $\frac{\beta^4}{\alpha^2}$___________.
$85$
$36$
$45$
$75$
Solution
$\frac{\mathrm{c}}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}}$ [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 $
$ \text { 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 $
