∠∠A = 2∠B，∠A=60

∴∠B=30，∠c = 90°

∴c=2b,a=3b

∴a？ =3b=b(b+c)

(2) Solution: The relationship a2=b(b+c) still holds.

Method 1: Prove: ∫∠ A = 2 ∠ B.

∴∠c= 180-∠a-∠b = 180-3∠b

Derived from sine theorem

a

Sina.com

=

b

Simbe

=

c

sinC

=2R

That is, a=2RsinA, b=2RsinB and c=2RsinC.

∴b(b+c)=2RsinB(2RsinB+2RsinC)

= 4r 2s inb[sin b+ sin( 180-3∠B)]

=4R2sinB(sinB+sin3∠B)

=4R2sinB(2sin2BcosB)

=4R2sin2B×sin2B

=4R2sin22B

∫a2 = 4r 2s in 2 a = 4r 2s in 22 b。

∴a2=b(b+c)