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March 25, 2017

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Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A + B).

  • Trig - ,

    make quick sketches of right-angled triangles for each of the two cases, and use Pythagoras to find the missing sides.

    first triangle:
    since sinA = 12/13 , y = 12 and r = 13
    x^2 + y^2 = r^2
    x^2 + 144 = 169
    x^2 = 25
    x = ±5, but we are in quad II, so x = -5
    then tanA = - 12/5

    2nd triangle:
    since tanB = -4/3

    tan(A-B) = (tanA - tanB)/(1 + tanAtanB)
    = (-12/5 - (-4/3) )/( 1 + (-12/5)(-4/3)
    = (-16/15) / (21/5)
    = -16/63

  • Trig - ,

    Are you sure it's negative?

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