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anonymous
 4 years ago
Let sin A = 12/13 with 90º≤A≤180º and tan B = 4/3 with 270º≤B≤360º. Find tan (A + B). I need help PLEASE
anonymous
 4 years ago
Let sin A = 12/13 with 90º≤A≤180º and tan B = 4/3 with 270º≤B≤360º. Find tan (A + B). I need help PLEASE

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myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0\[\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}=\frac{\sin(A)\cos(B)+\sin(B)\cos(A)}{\cos(A)\cos(B)\sin(A)\sin(B)}\] So we need to know the following: sin(A);sin(B);cos(A);cos(B) dw:1323414397246:dw

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0we have two right triangles: we need to use Pythagorean thm twice

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1323414566260:dw

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0sin(A)=12/13;sin(B)=3/5;cos(A)=5/13;cos(B)=4/5

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0so plug this into what I gave you above

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0\[\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}=\frac{\sin(A)\cos(B)+\sin(B)\cos(A)}{\cos(A)\cos(B)\sin(A)\sin(B)} \]
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