If $\sin heta=\sin 15^{\circ}+\sin 45^{\circ}$, where $0^{\circ}

Question

If $\sin 	heta=\sin 15^{\circ}+\sin 45^{\circ}$, where $0^{\circ} < 	heta < 180^{\circ}$, then $	heta=$

Aicharya · Accepted Answer

$\begin{aligned} 	ext { Given } \sin 	heta &=\sin 15^{\circ}+\sin 45^{\circ} \ &=2 \sin \left(\frac{15^{\circ}+45^{\circ}}{2}ight) \cos \left(\frac{15^{\circ}-45^{\circ}}{2}ight) \ &=2 \sin 30^{\circ} \cos 15^{\circ}=2 	imes \frac{1}{2} \cos 15^{\circ} \ \sin 	heta &=\cos 15^{\circ} \Rightarrow \sin 	heta=\sin \left(90^{\circ}-15^{\circ}ight) \Rightarrow 	heta=75^{\circ} \end{aligned}$

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

Solution