<PRE><B>Question 732211</B>
If sec a = 13/5 where a lies between 0° and 90° then the value of
2 sin a - 3 cos a / 4 sin a - 9 cos a is 
please send me step by step
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sec = 13/5 = r/x, so r = 13 and x = 5
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Find &quot;y&quot;:
x^2 + y^2 = r^2
y^2 = r^2-x^2
y = sqrt[13^2-5^2] = 12
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sin(a) = y/r = 12/13
cos(a) = x/r = 5/13
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2 sin a - 3 cos a / 4 sin a - 9 cos a
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 [2(12/13)-3(5/13)]/[4(12/13)-9(5/13)]
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= [9-3]/13
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= 6/13
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Cheers,
Stan H.
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