SOLUTION: What is the value of sin(A+B) if 5cosA minus 4 =0 and 12tanB+5=0?

Question 849575: What is the value of sin(A+B) if 5cosA minus 4 =0 and 12tanB+5=0?
Found 2 solutions by stanbon, lwsshak3:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
What is the value of sin(A+B) if 5cosA minus 4 =0 and 12tanB+5=0?
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cos(A) = 4/5
So, sin(A) = 3/5
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tan(B) = y/x = 5/12
So y = 5 and x = 12
Then r = sqrt(5^2+12^2) = 13
sin(B) = y/r = 5/13
cos(B) = 12/13
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Ans: So sin(A+B) = sin(A)cos(B) + cos(A)sin(B) = (3/5)(12/13)+(4/5)(5/13)
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= (36+20)/65
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= 56/65
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Cheers,
Stan H.
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Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
What is the value of sin(A+B) if 5cosA minus 4 =0 and 12tanB+5=0?
***
For:[0,π]
5cosA-4=0
cosA=4/5 ((3-4-5) right triangle)
sinA=3/5
..
12tanB+5=0
tanB=-5/12 ((5-12-13) right triangle in quadrant II)
sinB=5/13
cosB=-12/13
..
Identity: sin(A+B)=sinAcosB+cosAsinB
sin(A+B)=(3/5)*(-12/13)+(4/5)*(5/13)=-36/65+20/65=-16/65
..
calculator check:
cosA=4/5
A≈36.87˚
sinB=5/13
B≈157.38˚
A+B≈194.25
sin(A+B)≈sin(194.25)≈-0.2461..
Exact value=-16/65≈-0.2461..
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