Answer:
Given ;\[R=nH\]; so, \[\frac{{{u}^{2}}\sin
2\theta }{g}=n\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]
or\[\tan \text{
}\!\!\theta\!\!\text{ }\,\text{=}\frac{4}{n}\]or\[\theta ={{\tan }^{-1}}\left(
4/n \right)\]
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