Q. If $A = 130{^\circ}$ and $x = \sin{A + \cos A}$, then
A) $x > 0$
B) $x < 0$
C) $x = 0$
D) $x \geq 0$
Explanation
(a)
$x = \sin{130{^\circ} + \cos{130{^\circ}}}$
$= \sin{50{^\circ} - \sin{40{^\circ} > 0\ \ \ \ }}$ $\left\lbrack \because\sin{x\ is\ increasing\ for\ 0 < x < \frac{\pi}{2}} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.