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For reference: In the figure, AOB is a quadrant and the quadrilaterals OMNL and LTQK are square.

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My progress..I would like a solution by geometry...by trigonometry it is solved: $\triangle QOL: \frac{r}{\sin45}=\frac{r\sqrt2}{2\sin 2\theta} \implies \sin2\theta =\frac{1}{2} \therefore \theta = 15^\circ$

enter image description here

ACB
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peta arantes
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1 Answers1

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Flip squares vertically.

image

$\angle BDC=\angle BEC=x$ (angles of same segment)

$\triangle BFE$ and $\triangle BGA$ are congruent.

Therefore, $\angle BEF=\angle BAG=x$

$\angle CAG=45^\circ=2x+x$
$\implies x=15^\circ$

ACB
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