Question

If $a=\sin {170}^{\circ }+\cos {170}^{\circ }$, then correct option is

• A. $a>0$
• B. $a<0$
• C. $a=0$
• D. $a=1$

Answer 1

The correct option is B $a<0$

$\sin {170}^{\circ }=\sin (180-10{)}^{\circ }=\sin {10}^{\circ }$
$\cos {170}^{\circ }=\cos (180-10{)}^{\circ }=-\cos {10}^{\circ }$

$a=\sin {170}^{\circ }+\cos {170}^{\circ }$
$=\sin {10}^{\circ }-\cos {10}^{\circ }$


From graph we can observed that,
For $x\in [0,{45}^{\circ }),\cos x>\sin x$
So, $\cos {10}^{\circ }>\sin {10}^{\circ }$
$\therefore a=\sin {10}^{\circ }-\cos {10}^{\circ }<0$