Let ABC be a triangle in which AB=BC. Let X b | vedclass

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(B) Given in $	riangle ABC$,$AB=BC$ and $\frac{AX}{XB} = \frac{AB}{AX} = k$. Since $AB = AX + XB$,we have $\frac{AB}{AX} = \frac{AX+XB}{AX} = 1 + \fr

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) Given in $	riangle ABC$,$AB=BC$ and $\frac{AX}{XB} = \frac{AB}{AX} = k$. Since $AB = AX + XB$,we have $\frac{AB}{AX} = \frac{AX+XB}{AX} = 1 + \frac{XB}{AX} = 1 + \frac{1}{k}$. Thus,$k = 1 + \frac{1}{k} \Rightarrow k^2 - k - 1 = 0$. Since $k > 0$,$k = \frac{1+\sqrt{5}}{2}$. Then $\frac{AX}{AB} = \frac{1}{k} = \frac{2}{1+\sqrt{5}} = \frac{\sqrt{5}-1}{2}$. Let $\angle ABC = 	heta$. Since $AB=BC$,$\angle BAC = \angle BCA = \frac{180^{\circ}-	heta}{2} = 90^{\circ} - \frac{	heta}{2}$. In $	riangle AXC$,$AC=AX$,so $\angle AXC = \angle ACX = \angle BCA = 90^{\circ} - \frac{	heta}{2}$. Then $\angle XAC = 180^{\circ} - 2(90^{\circ} - \frac{	heta}{2}) = 	heta$. Since $\angle BAC = 90^{\circ} - \frac{	heta}{2}$,we have $\angle BAX = \angle BAC - \angle XAC = (90^{\circ} - \frac{	heta}{2}) - 	heta = 90^{\circ} - \frac{3	heta}{2}$. In $	riangle BAX$,by the Law of Sines: $\frac{AX}{\sin 	heta} = \frac{AB}{\sin(90^{\circ}-\frac{	heta}{2})}$. $\frac{AX}{AB} = \frac{\sin 	heta}{\cos(	heta/2)} = \frac{2\sin(	heta/2)\cos(	heta/2)}{\cos(	heta/2)} = 2\sin(	heta/2)$. Equating the two expressions for $\frac{AX}{AB}$: $2\sin(	heta/2) = \frac{\sqrt{5}-1}{2} \Rightarrow \sin(	heta/2) = \frac{\sqrt{5}-1}{4} = \sin 18^{\circ}$. Therefore,$\frac{	heta}{2} = 18^{\circ} \Rightarrow 	heta = 36^{\circ}$.

Let $ABC$ be a triangle in which $AB=BC$. Let $X$ be a point on $AB$ such that $AX:XB=AB:AX$. If $AC=AX$,then the measure of $\angle ABC$ equals (in $^{\circ}$)

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