Question

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

• A. ${18}^{\circ }$
• B. ${36}^{\circ }$
• C. ${54}^{\circ }$
• D. ${72}^{\circ }$

Answer 1

The correct option is B ${36}^{\circ }$

From the data,
$AB=c,AC=b,BC=a,\angle ABC=B$
$AX:XB=AB:AX\&AC=AX$

$A{X}^2=AB\times XB\Rightarrow b^2=c(c-b)\Rightarrow b^2+bc-c^2=0\Rightarrow \frac{b}{c}=\frac{\sqrt{5}-1}{2}$

Now,
$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{2c^2-b^2}{2c^2}=1-\frac{b^2}{2c^2}=\frac{\sqrt{5}+1}{4}$

$\Rightarrow B={36}^{\circ }$