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If $x = \sin 130 ° + \cos 130 °$ then

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x > 0

x \geq 0

x = 0

x < 0

answer is C.

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Detailed Solution

x = \sqrt{2} [\frac{1}{\sqrt{2}} \sin 130 + \frac{1}{\sqrt{2}} \cos 130]

$= \sqrt{2} [\sin 130. \cos 45 + \sin 45. \cos 130]$

$= \sqrt{2} [\sin (130 + 45)]$

$= \sqrt{2} \sin 175 = \sqrt{2} \sin (180 - 5)$

$= \sqrt{2} \sin 5$

$x > 0$

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