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

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(B) Given: $x = \sin 130^\circ \cos 80^\circ$ and $y = \sin 80^\circ \cos 130^\circ$.Using the identity $\sin(180^\circ - 	heta) = \sin 	heta$ and $

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given: $x = \sin 130^\circ \cos 80^\circ$ and $y = \sin 80^\circ \cos 130^\circ$.Using the identity $\sin(180^\circ - heta) = \sin heta$ and $\cos(180^\circ - heta) = -\cos heta$:$x = \sin(180^\circ - 50^\circ) \cos 80^\circ = \sin 50^\circ \cos 80^\circ$. Since $50^\circ$ and $80^\circ$ are in the first quadrant,$\sin 50^\circ > 0$ and $\cos 80^\circ > 0$,so $x > 0$.$y = \sin 80^\circ \cos(180^\circ - 50^\circ) = \sin 80^\circ (-\cos 50^\circ)$. Since $\sin 80^\circ > 0$ and $\cos 50^\circ > 0$,$y Now,$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)$.Using $2 \sin heta \cos heta = \sin 2 heta$,we get $xy = (\frac{1}{2} \sin 260^\circ)(\frac{1}{2} \sin 160^\circ) = \frac{1}{4} \sin(270^\circ - 10^\circ) \sin(180^\circ - 20^\circ) = \frac{1}{4} (-\cos 10^\circ)(\sin 20^\circ)$.Since $\cos 10^\circ > 0$ and $\sin 20^\circ > 0$,$xy Since $xy$ is a small negative value,$z = 1 + xy$ implies $0 < z < 1$.

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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