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if tan a = -5/12 and cos a> 0, find sin a
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|dw:1396930995255:dw| tan(a) = sin(a) / cos(a) = -5/12 Since cos(a) > 0, sin(a) must be < 0.
First find AC using Pythagoras Theorem. sin(a) = BC / AC = 5/? Put a negative sign in front of it to make sin(a) < 0.
How do I do that I don't know how
Pythagoras theorem: AC^2 = AB^2 + BC^2 = 12^2 + 5^2 = 144 + 25 = 169 AC = sqrt(169) = 13 |sin(a)| = BC/AC = 5/13 Put a negative sign to make sin(a) negative. sin(a) = -5/13
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