Recent content by pokemonDoom

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    How do you compute trig ratios for angles greater than 90 degrees?

    ... I get it ! But isn't the angle opposite side DC <DOC ... ? I am not trying to sound dumb but what youre trying to say is exactly what the question is . Why should sineAOC =CD/OC , when its perfectly legible that sineDOC = CD/OC . In essence what youre trying to say is for any...
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    How do you compute trig ratios for angles greater than 90 degrees?

    Hello Tim , Im confused ... how can the cosine of <AOC be AD/OC ... if I know better , its probably a typo . and in triangle-DOC (which lies squarely on the second quadrant of my graph) how is it that the sine of <AOC is equivalent to the sine of <DOC ... This is the question in the...
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    How do you compute trig ratios for angles greater than 90 degrees?

    Heres my problem ... For the triangle in the second quadrant , why does my book state sin(90+theta) = a/r and cos(90+theta) = -b/r when clearly (90+theta) isn't even one of the angles in the triangle ... ? This was what I meant when I said why not calculate (180-90-theta) ... I...
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    How do you compute trig ratios for angles greater than 90 degrees?

    hmm ... but that doesn't solve my problem ... I need to know why the so called "reference angle" or the "corresponding acute angle" actually works for angles greater than 90 degrees , when taking sines and cosines or tangents . Is there a geometric argument for it ?
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    How do you compute trig ratios for angles greater than 90 degrees?

    Say I have an acute angle A in standard position in the first quadrant with a rotating arm of length r , terminating at coordinate P(a,b) . Now say I rotate it 90 degrees further from that position to the second quadrant , this ends up at coordinate Q(-a,b) . I draw perpendiculars from those...
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