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pre-calculus

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If sin a = 3/5 when Pi/2<a<Pi and
cos B = -12/13 when Pi<B<3Pi/2, then what is the value of sin(a+b)?

I don't get the question... Help appreciated.

  • pre-calculus -

    make two separate diagrams,
    one will be the 3,4,5 right-angled triangle,
    the other will be the 5,12,13 right-angled triangle

    given sin a = 3/5 and a is in the II quadrant, so cos a = -4/5

    given cos B = -12/13 and B is in III, so
    sin B = - 5/13

    now you have to know the expansion for
    sin(a+b) = sina cosb + cosa sinb
    =(3/5)(-12/13) + (-4/5)(-5/13) = -16/65

  • pre-calculus -

    Thanks!

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