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From the below figure

tan(a)=1/(2-tan(15)

tan(15)=tan(45-30)=(1-1/3^0.5)/(1+1/3^0.5)=2-3^0.5

tan(a)=1/(2-2+3^0.5)

hence a =30 degress
ans D


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We have ∠ECF = 90-15 = 75° = ∠EFC => EF = EC = DE = DF/2
We have ∠HEC = 30° => HC = EC/2

We have DF*HC = DC*FC (both equal two times the area of the triangle DFC)
=> 2*EC*EC/2 = BC*FC
=> EC^2 = BC*FC
=> FC/EC = EC/BC
=> triangle ECF and triangle BCE are similar
=> ∠EBC = ∠FEC = 30°

The answer is D
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Refer to the image attached:

I have copied the triangle on the side CD to side BC.
Now the triangle CDE and BCP are congruent.

Angle EDC = Angle PCB
Angle ECD = Angle PBC
Angle CED = Angle BPC
Since CD = BC ( from square property),
We know now,

DE = CP
EC = PB

ECP = BCD – ECD – BCP
ECP = 90° – 15° – 15°
ECP = 60°

Because, we have an isosceles triangle with CP = CE, and DCE = BCF = 15° already and now with ECF = 60°, the other two sides of triangle ECP will also be 60° making it a equilateral triangle.

Since sum of all angles of the vertices of a triangle is 180°,

we can now say BPC = 150°.

Also, since sum of all the 3 angles around point P, will be 360°,
we can also say BPE = 150°.

Therefore the triangles EPB and BPC are congruent with 2 sides are same and one angle same ( with EP = CP, EPB = BPC, and PB is one of the side for each of the triangles).

Thus EBP = CBP = 15°.

Therefore, EBC = EBP + CBP = 15° + 15° = 30°.

So the correct answer choice is D
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