If sin theta = sin 15^{circ} + sin 45^{circ}, | vedclass

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(A) Given $\sin 	heta = \sin 15^{\circ} + \sin 45^{\circ}$.Using the formula $\sin C + \sin D = 2 \sin \left( \frac{C+D}{2} ight) \cos \left( \frac

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$75$

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(A) Given $\sin 	heta = \sin 15^{\circ} + \sin 45^{\circ}$.Using the formula $\sin C + \sin D = 2 \sin \left( \frac{C+D}{2} ight) \cos \left( \frac{C-D}{2} ight)$:$\sin 	heta = 2 \sin \left( \frac{15^{\circ} + 45^{\circ}}{2} ight) \cos \left( \frac{15^{\circ} - 45^{\circ}}{2} ight)$$\sin 	heta = 2 \sin 30^{\circ} \cos(-15^{\circ})$Since $\cos(-x) = \cos x$ and $\sin 30^{\circ} = \frac{1}{2}$:$\sin 	heta = 2 	imes \frac{1}{2} 	imes \cos 15^{\circ}$$\sin 	heta = \cos 15^{\circ}$Since $\cos 15^{\circ} = \sin(90^{\circ} - 15^{\circ}) = \sin 75^{\circ}$:$\sin 	heta = \sin 75^{\circ}$Therefore,$	heta = 75^{\circ}$.

If $\sin \theta = \sin 15^{\circ} + \sin 45^{\circ}$,where $0^{\circ} < \theta < 180^{\circ}$,then $\theta =$ (in $^{\circ}$)

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