Answer:
Given \[{{R}_{A}}={{R}_{B}}\] or \[\frac{{{(\sqrt{2\upsilon })}^{2}}\sin (2\times {{15}^{\circ }})}{g}=\frac{{{\upsilon }^{2}}\sin 2\theta }{g}\] or \[\sin 2\theta =1=\sin {{90}^{\circ }}\] \[\theta ={{45}^{\circ }}\]
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