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    Hi. I get how to do this question, it's just that I'm stuck on a little part of it, which ruins my answer. Here goes:

    Given that tanA = \frac{7}{24} where A is reflex, and sinB = \frac{5}{13} where B obtuse, calculate the value of sin(A + B).

    From this, I can conclude that 180 \leq A \leq 360 and that  90 \leq B \leq 180 .

    sin(A + B) = sinAcosB + cosAsinB

    cos^2 B = 1 - (\frac{5}{13})^2
    cosB = \frac{-12}{13}
    *** Am I right here to say it is negative as B is in the second quadrant and so cosine is negative.

     1 + tan^2 A = sec^2 A
     secA = \pm \frac{25}{24}
     cosA = \frac{-24}{25}
    ***This is where I'm stuck - Since it is a reflex angle, it covers both quadrants 3 and 4. So how do I know if it positive or negative as the third quadrant is negative for cosine but the fourth is positive. Hope I'm making sense here!?
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    The fact that A is a reflex angle and its tan is positive is sufficient to narrow it down to one quadrant. Can you work it out from that?
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    (Original post by ghostwalker)
    The fact that A is a reflex angle and its tan is positive is sufficient to narrow it down to one quadrant. Can you work it out from that?
    Oh thanks! :facepalm:
    So we're talking about the third quadrant here, which means cosine is negative.
    Thanks again.
 
 
 
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Updated: February 14, 2010
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