Maybe simplest is to solve for in the diagram.
If we use the corners of the square as circle centres, with radius "x",
and using the fact that the angle at the centre of a circle is twice the angle at the circumference
for angles standing on a common arc,
then it follows that if angle ACE is 30, then the angles ADE and AED are 15 degrees.
Triangle AGC is equilateral as all sides are R.
The angles of AGC are all , so
Your problem was poorly worded,corrected by diagram but failing to redefine the. givens
At C draw an arc r=1 from Dto B and at B an arc fromCto E.These arcs meet at A along the perpendicular bisector of CB and DE
DA is a chord in a quarter circle. EDA is given 15 DE is tangent to circle at D
Arc DA = 30 Angle DCA =30 Angle ACB=60
Which picture is not correct?
What is wrong in my statement?
I have drawn PAQ parallel to DME. EMAQ is a rectangle. Angle AEM = 15 degrees is given in the problem. In the rectangle EMAQ angle MEA = Angle EMQ = 15 degrees. R is the point of intersection of the sector and MA. I have drawn SRT parallel to EMD.