Question

If $x=\sin 130°\cos 80°,y=\sin 80°\cos 130°,z=1+xy$ which one of the following is true?

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

Answer 1

The correct option is B

$x>0,y<0,0<z<1$

Explanation for the correct option.

Step 1: Find $x$.

$\begin{array}{rcl}x& =& \sin 130°\cos 80°\\ & =& \sin \left(90°+40°\right)\cos 80°\left[by\sin \left(90°+\theta \right)=\cos \theta \right]\\ & =& \cos 40°\cos 80°\end{array}$

That means $x>0$.

Step 2: Find $y$.

$\begin{array}{rcl}y& =& \sin 80°\cos 130°\\ & =& \sin 80°\cos \left(90°+40°\right)\left[by\cos \left(90°+\theta \right)=-\sin \theta \right]\\ & =& -\sin 80°\sin 40°\end{array}$

That means $y<0$.

Step 3: Find $z$.

$z=1+xy$

As $x>0$, and $y<0$, so $xy<0$.

So, $z=1+xy<1$.

$\Rightarrow 0<z<1$

Therefore, $x>0$, $y<0$, and $0<z<1$.

Hence, option B is correct.