If x = sin 130^circ cos 80^circ,y = sin 80^ci | vedclass

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Question

(B) Given $x = \sin 130^\circ \cos 80^\circ$ and $y = \sin 80^\circ \cos 130^\circ$.Since $130^\circ$ is in the second quadrant,$\sin 130^\circ = \sin

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $x = \sin 130^\circ \cos 80^\circ$ and $y = \sin 80^\circ \cos 130^\circ$.Since $130^\circ$ is in the second quadrant,$\sin 130^\circ = \sin(180^\circ - 50^\circ) = \sin 50^\circ = \cos 40^\circ > 0$ and $\cos 130^\circ = -\sin 40^\circ Also,$\cos 80^\circ > 0$ and $\sin 80^\circ > 0$.Therefore,$x = (\sin 130^\circ)(\cos 80^\circ) = ( ext{positive})( ext{positive}) > 0$.And $y = (\sin 80^\circ)(\cos 130^\circ) = ( ext{positive})( ext{negative}) Since $x > 0$ and $y Given $z = 1 + xy$,since $xy Also,$xy = (\sin 130^\circ \cos 80^\circ)(\sin 80^\circ \cos 130^\circ) = (\sin 130^\circ \cos 130^\circ)(\sin 80^\circ \cos 80^\circ) = \frac{1}{2} \sin 260^\circ \cdot \frac{1}{2} \sin 160^\circ = \frac{1}{4} (-\sin 80^\circ)(\sin 20^\circ) = -\frac{1}{4} \sin 80^\circ \sin 20^\circ$.Since $\sin 80^\circ \sin 20^\circ$ is a small positive value,$xy$ is a small negative value,so $0 < z < 1$.

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

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