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Bunuel
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I am not able to delete this post. Please refer the post below.

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Let length of the side of the square be a
so the radius of the quarter circle is also a
let radius of the smaller circle be r

since OAP=45 degree as it is the angle the diagonal of the square makes with the side,
so \(OA=\sqrt{2}r\)

as \(tan45=\frac{OP}{OA} (OP=r)\)

Let QC be m

so in triangle OCR, \(sin OCR=\frac{OR}{OC} =\frac{r}{(r+m)}\)
lets find out m in terms of r,
\(AO+OQ+QC=AC\)
i.e. \( \sqrt{2}r +r+m =\sqrt{2}a ----------(1)\)

AO+OQ =AQ
i.e.\( \sqrt{2}r +r = a --------- (2)\\
\)
Substituting equation 2 in 1 we get m=r
so \(sin OCR=\frac{r}{2r}=\frac{1}{2}\)
OCR=30 degree
and the angle in question is 45-30 =15
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chetan sir, please spread light on this Buneul-like dark

As GMAT does not test tang-cos there must be another simpler explanation
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Hi,

I found this quite challenging, but maybe it's just me. I was surprised 50% got it correct, or maybe, I just did it too complicated, but here is my way, just using basic knowledge on squares and the pythagorean.

Before, we have to prove co-linearity of lines AC and AP, this can be done and is not too complicated, so just assume it has already been proven and check the following:
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