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Sbdenney

  • 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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  1. Sbdenney
    • 4 years ago
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    I need the fraction

  2. Sbdenney
    • 4 years ago
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    ? PLEASE!

  3. myininaya
    • 4 years ago
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    \[\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}=\frac{\sin(A)\cos(B)+\sin(B)\cos(A)}{\cos(A)\cos(B)-\sin(A)\sin(B)}\] So we need to know the following: sin(A);sin(B);cos(A);cos(B) |dw:1323414397246:dw|

  4. myininaya
    • 4 years ago
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    we have two right triangles: we need to use Pythagorean thm twice

  5. myininaya
    • 4 years ago
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    |dw:1323414566260:dw|

  6. myininaya
    • 4 years ago
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    sin(A)=12/13;sin(B)=-3/5;cos(A)=-5/13;cos(B)=4/5

  7. myininaya
    • 4 years ago
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    so plug this into what I gave you above

  8. myininaya
    • 4 years ago
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    \[\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}=\frac{\sin(A)\cos(B)+\sin(B)\cos(A)}{\cos(A)\cos(B)-\sin(A)\sin(B)} \]

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