If $A = 130^{0}$ and $x = sin A + cos A$ , then

x > 0

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x < 0

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x \geq 0

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x \leq 0

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Solution

The correct option is B $x > 0$
Given $A = 130^{o}$

$x = sin A + cos A = \sqrt{2} sin (\frac{π}{4} + A) a s [sin (A + B = sin A cos B + sin B cos A)]$

$= \sqrt{2} sin (45^{o} + 130^{o})$

$= \sqrt{(} 2) sin (175^{o})$

$∴ sin (175^{o}) > 0$

$x > 0$

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