Question

If $x=sin{130}^{\circ }cos{80}^{\circ },y=sin{80}^{\circ }cos{130}^{\circ },z=1+xy,$
which one of the following is true

• A. x > 0, y > 0, z > 0
• B. x > 0, y < 0, 0 < z < 1
• C. x > 0, y < 0, z > 1
• D. x < 0, y < 0, 0 < z < 1

Answer 1

The correct option is B x > 0, y < 0, 0 < z < 1

$x=sin{130}^{\circ }cos{80}^{\circ },y=sin{80}^{\circ }cos{130}^{\circ }$
$\Rightarrow x=cos{40}^{\circ }cos{80}^{\circ },y=-sin{80}^{\circ }sin{40}^{\circ }$
So, x > 0 and y < 0 and xy < 0
Now, z = 1 + xy $\Rightarrow$ 0 < z < 1.