In the triangle ABC, BC = B-A and B-A = 80 ° then the cosine of the inner angle c is

In the triangle ABC, BC = B-A and B-A = 80 ° then the cosine of the inner angle c is

∵bc=b2-a2,∴sinBsinC=sin2B-sin2A,∴sinBsinC=（sinB+sinA）（sinB-sinA）,∴sinBsinC=4sinB+A 2 cosB−A 2 cosB+A 2 sinB−A 2 =sinCsin（B-A）,∴sinB=sin（B-A）,∴2B-A=180°,∵B-A=80°,∴B=100°,...

In the triangle ABC, BC = B ^ 2-A ^ 2, and B-A = 80, then the cosine value of internal angle c is

∵bc=b2-a2,∴sinBsinC=sin2B-sin2A,∴sinBsinC=（sinB+sinA）（sinB-sinA）,∴sinBsinC=4sin
 
B+A2
cos
B−A2
cos
B+A2
sin
B−A2
=sinCsin（B-A）,∴sinB=sin（B-A）,∴2B-A=180°,∵B-A=80°,∴B=100°,A=20°,∴C=60°,∴cosC=
twelve
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