Question

锐角三角形ABC中,sinA=5分之根号5,sinB=10分之根号10.

Answer 1

1)cosA=√(1-sin²A)=2√5/5

cosB=√(1-sin²B)=3√10/10

cosC=-cos(A+B)

=sinAsinB-cosAcosB

=√5/5*√10/10-2√5/5*3√10/10

=-√2/2

C=135°

2)a/b=sinA/sinB=√2

a=√2b

a-b=√2-1

a=√2,b=1

c=asinC/sinA=√5