please help me!

both question rely on a concept
sin (-A) = -sin A
cos (-A) = cos A

1
remind that A+B+C = 180
sin B = sqrt(1-(cosB)^2 ) = 3/5
cos A = sqrt(1-(sinA)^2 ) = 8/13

cos C
= cos (180 - A - B)
= -cos (A+B)
= -cosAcosB + sinAsinB
= -(8/13)(4/5) + (5/13)(3/5)
= -17 / 65

2006-10-31 13:56:39 補充：
2 , sin A = 9/41cos A = sqrt(1-(sinA)^2 ) = 40/41cos C = cos (180-(A B)) = -cos (A B) = -7/25sin C = sqrt(1-(cosC)^2 ) = 24/25sin B= sin (180 -(A C))= sin (A C)= sinAcosC sinCcosA= -63/1025 960/1025= 897 / 1025

2006-10-31 14:01:46 補充：
mistake in question 1sin A = 5/13 .... cos A = 12/13cosB = 4/5 .... sin B = 3/5the method is correct , but the answer should be cos C = 15/65 - 48/65 = -33/65

Is it ask you to answer without calculator?

2006-10-31 14:24:46 補充：
1)cosA=12/13cosC=cos(180-(A B))=cos(A B)=cosAcosB-sinAsinB=48/65-3/13=33/652)tanA=9/40tan(A B)=24/7tanA tanB/1-tanAtanB=24/79/40 tanB=24/7-27/35tanB62/35tanB=897/280tanB=897/496Since 開方(496^2 897^2) = 1025sinB=897/1025.