∵cosB=5/13,
∴sinB=√1-cos^2B
=12/13
∵sinB>sinA,
所以B>A,故A是锐角,cosA=√3/2
所以cosC = cos(π-A-B) = -cos(A+B)
=sinAsinB - cosAcosB
=1/2 * 12/13 - √3/2 * 5/13
=(12-5√3)/26
∵cosB=5/13,
∴sinB=√1-cos^2B
=12/13
∵sinB>sinA,
所以B>A,故A是锐角,cosA=√3/2
所以cosC = cos(π-A-B) = -cos(A+B)
=sinAsinB - cosAcosB
=1/2 * 12/13 - √3/2 * 5/13
=(12-5√3)/26