If A = 130^circ and x = sin A + cos A,then | vedclass

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(A) Given $A = 130^\circ$. $x = \sin 130^\circ + \cos 130^\circ$. Using the identity $\sin 	heta = \cos(90^\circ - 	heta)$,we have $\sin 130^\circ =

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$x > 0$

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(A) Given $A = 130^\circ$. $x = \sin 130^\circ + \cos 130^\circ$. Using the identity $\sin 	heta = \cos(90^\circ - 	heta)$,we have $\sin 130^\circ = \cos(90^\circ - 130^\circ) = \cos(-40^\circ) = \cos 40^\circ$. So,$x = \cos 40^\circ + \cos 130^\circ$. Using the sum-to-product formula $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$: $x = 2 \cos(\frac{40^\circ + 130^\circ}{2}) \cos(\frac{40^\circ - 130^\circ}{2}) = 2 \cos(85^\circ) \cos(-45^\circ) = 2 \cos 85^\circ \cos 45^\circ$. Since $85^\circ$ is in the first quadrant,$\cos 85^\circ > 0$. Since $45^\circ$ is in the first quadrant,$\cos 45^\circ > 0$. Therefore,$x = 2 \cos 85^\circ \cos 45^\circ > 0$.

If $A = 130^\circ$ and $x = \sin A + \cos A$,then

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