If cos x = 12/13 and x is an acute angle, how do you find the value of cos (180 - x)?

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Answer 1 · 2016-09-24T20:24:38

$cos (180^{\circ} - x) = - \frac{12}{13}$ $cos(180^@-x)=-12/13$

We will use the cosine angle subtraction formula:

$cos (a - b) = cos (a) cos (b) + sin (a) sin (b)$ $cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$

Thus:

$cos (180^{\circ} - x) = cos (180^{\circ}) cos (x) + sin (180^{\circ}) sin (x)$ $cos(180^@-x)=cos(180^@)cos(x)+sin(180^@)sin(x)$

Note that $cos (180^{\circ}) = - 1$ $cos(180^@)=-1$ and $sin (180^{\circ}) = 0$ $sin(180^@)=0$ :

$cos (180^{\circ} - x) = (- 1) cos (x) + (0) sin (x)$ $cos(180^@-x)=(-1)cos(x)+(0)sin(x)$

$cos (180^{\circ} - x) = - cos (x)$ $cos(180^@-x)=-cos(x)$

And since in this scenario $cos (x) = \frac{12}{13}$ $cos(x)=12/13$ :

$cos (180^{\circ} - x) = - \frac{12}{13}$ $cos(180^@-x)=-12/13$