In a triangle $ABC,$if $\angle A=30^\circ,b=10$ and $a=x$,then the values of $x$ for which there are $2$ possible triangles is In a triangle $ABC,$if $\angle A=30^\circ,b=10$ and $a=x$,then the values of $x$ for which there are $2$ possible triangles is given by
$(A)5<x<10(B)x<\frac{5}{2}(C)\frac{5}{3}<x<10(D)\frac{5}{2}<x<10$

$\cos\angle A=\frac{AB^2+AC^2-BC^2}{2\times AB\times AC}=\frac{AB^2+100-x^2}{20\times AB}$
$\frac{\sqrt3}{2}=\frac{AB^2+100-x^2}{20\times AB}$
I am stuck here.


$$x^2=c^2+100-20c\frac {\sqrt 3}2$$
$$c^2-10\sqrt3 c+100-x^2=0$$
Two triangles exist --  equation has two roots:
$$\frac D4=75-(100-x^2)>0$$
$$x^2> 25$$
$$x > 5$$
$$x<b+c=10+c_1 \Rightarrow x<10$$
