Question

If $x=\sin {130}^o+\cos {130}^o$ then

• A. $x<0$
• B. $x=0$
• C. $x>0$
• D. $x\ge 0$

Answer 1

The correct option is C $x>0$

$x=\sin {130}^{\circ }+\cos {130}^{\circ }$

$⟹x=\sin ({180}^{\circ }-{50}^{\circ })+\cos ({180}^{\circ }-{50}^{\circ })=\sin {50}^{\circ }-\cos {50}^{\circ }$

For ${45}^{\circ }<\theta <{90}^{\circ },\sin \theta >\cos \theta$

So $\sin {50}^{\circ }>\cos {50}^{\circ }⟹\sin {50}^{\circ }-\cos {50}^{\circ }>0⟹x>0$