# Solving the equation $\sin 40^{\circ}=\cos x$

How to solve this equation (finding $x$ ):

$$\sin 40^\circ=\cos x$$

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HINT $$\sin(y^{\circ}) = \cos(90^{\circ}-y^{\circ})$$

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Typo: you typed hint instead of answer. –  Git Gud Apr 11 '13 at 16:54
@GitGud: He does that often.;) –  Inceptio Apr 11 '13 at 17:01
@Marvis why such a tiny joke make you take such major decisions? Don't be insulted so badly by two users of the whole site :) –  Ruslan Apr 11 '13 at 17:22
@GitGud I think you need to relearn your English. –  user17762 Apr 12 '13 at 0:11
@Inceptio: You have been on this site for just $50$ days and I believe you do not have enough sample data to make this statement. Given the OP's level, I believe this is an appropriate hint. –  user17762 Apr 12 '13 at 0:11

Hint:

$\sin \angle BAC= \dfrac{BC}{AC}$ and $\sin\angle BAC=\cos(90-\angle BAC) =\dfrac{AB}{AC}$. I will let the figure to do the further talking.

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sin(40)=coS(50) because siny = coS(90-y)

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