Determine $\tan(α)$ in the figure below

Question

In the figure below $ABCD$ is a square. Determine $\tan(α)$.

Alternatives: $A)\, 0.9\\ B)\, 0.8\\ C)\, 0.7\\ D)\, 0.6\\ E)\, 0.5$

$FG=r\\ AB=BC=CD+AD = l\\ \triangle AHG: AH^2=(l+r)^2-(l-r)^2 \implies AH^2=4lr\\ \therefore AH = 2\sqrt{lr}$ $\tan \alpha = \frac{r}{CF}=\frac{r}{l-2\sqrt{lr}}$

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Answer 1

According to your work, we have $AH = 2 \sqrt r$ if we let $l = 1$.

Then, $CF = 1 – 2 \sqrt r$.

From $\triangle EGI$ [should be $ \triangle EFG$], we get $(0.5 – r)^2 = r^2 + (2 \sqrt r – 0.5)^2$

Then, $r = 0.16$. Result follows.