In an extracurricular activity, Liu Wei made two right-angled triangles out of cardboard, as shown in Figures ① and ②. In Figure 1, ∠ B = 90, ∠ A = 30, BC=6.

Solution: (1) becomes smaller; (2) Question ①: ∵∠ B = 90, ∠ A = 30, BC=6.

∴AC= 12

∠∠FDE = 90 degrees, ∠DEF=45 degrees, DE=4 degrees.

∴DF=4

Connect FC and set FC∨AB.

∴∠FCD=∠A=30

∴△FDC in Rt, DC=

∴AD=AC-DC= 12-

∴AD=( 12- )cm，fc∨ab；

Question 2: let AD=x,

In Rt△FDC, fc 2 = DC 2+FD 2 = (12-x) 2+16.

(i) When FC is the hypotenuse,

From AD 2 +BC 2 =FC 2,

x 2 +6 2 =( 12-x) 2 + 16，

x =；

(II) When AD is the hypotenuse,

From FC 2 +BC 2 =AD 2,

( 12-x) 2 + 16+6 2 =x 2，

X = > 8 (irrelevant);

(III) When BC is the hypotenuse,

From AD 2 +FC 2 =BC 2,

x 2 +( 12-x) 2 + 16=36，x 2 - 12x+62=0，

∴ Equation has no solution,

∴ From (i), (II) and (III), when x= cm, the triangle with three sides as the length of line segments AD, FC and BC is a right triangle;

Question 3:

There is no position ∠ fcd = 15 for the following reasons:

Suppose ∠ fcd = 15.

∠∠EFC = 30

Make the bisector of ∠EFC and cross AC at point P.

Then ∠ EFP =∠ CFP = 15, ∠ DFE+∠ EFP = 60.

∴PD=，PC=PF=2FD=8

∴PC+PD=8+ > 12

∴ There is no such position as ∠ fcd = 15.