( 1)cos∠ADC = 1/7 ,sin∠adc=√ 1-(cos∠adc)^2]=4√3/7

sin∠ADB=sin∠ADC=4√3/7

cos∠ADB=-cos∠ADC=- 1/7

sin∠BAD = sin( 180-∠ABD-∠ADB)= sin(2π/3-∠ADB)=√3/2 cos∠ADB+ 1/2 sin∠ADB

=√3/2*(- 1/7)+ 1/2*4√3/7

=3√3/ 14

So sin ∠ bad =1114.

(2)AB/sin∠ADB=BD/sin∠BAD

BD =(AB * sin∠BAD)/sin∠ADB =(8 * 3√3/ 14)/(4√3/7)= 3

BD=3

(3)BC=BD+CD=3+2=5

ac^2=ab^2+bc^2-2ab*bccosπ/3=8^2+5^2-2*8*5=9

AC=3

Namely: sin ∠ bad =114, BD = 5, AC = 3.

two

( 1):(b-2a)cosC+ccosB=0

From sine theorem:

(sinB-2sinA)cosC+sinCcosB=0

sinCcosB+sinBcosC-2sinAcosC=0

sin(C+B)-2sinAcosC=0

sinA-2sinAcosC=0

Sina is not 0.

1-2 cos = 0

cosC= 1/2

C=π/3

(2)c^2=a^2+b^2-2abcosπ/3

a^2+b^2-ab=c^2

a^2+(3a)^2-3a*a=(√7)^2

A= 1 or a=- 1 (omitted)

a= 1，b=3a=3

s = 1/2 absinπ/3 = 1/2 * 1 * 3 *√3/2 = 3 √/ 4

S=3√/4

three

2asinA=(2b+c)sinB+(2c+b)sinC

From sine theorem:

2a^2=b(2b+c)+c(2c+b)

2a^2=2b^2+bc+2c^2+bc

a^2=b^2+c^2+bc

b^2+c^2-a^2=-bc

cosa=(b^2+c^2-a^2)/2bc=-bc/2bc=- 1/2

cosA=- 1/2

A=2π/3