Question 916112: Given sinA:4/5, cosB=5/13 both A and B in quadrant I, find tan(A+B)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Given sinA:4/5, cosB=5/13 both A and B in quadrant I, find tan(A+B)
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sinA=4/5
cosA=3/5(working with a 3-4-5 reference right triangle)
..
cosB=5/13
sinB=12/13(working with a 5-12-13 reference right triangle)
..
sin(A+B)=sinAcosB+cosAsinB=4/5*5/13+3/5*12/13=20/65+36/65=56/65
cos(A+B)=cosAcosB-sinAsinB=3/5*5/13-4/5*12/13=15/65-48/65=-33/65
tan(A+B)=sin(A+B)/cos(A+B)=-56/33
..
Check:
sinA=4/5
A=53.13˚
cosB=5/13
B=67.38˚
A+B=53.13+67.38=120.51˚
tan(A+B)=tan(120.51˚)=-1.6969...
as calculated: tan(A+B)=-56/33=-1.6969...
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