A) \[{{30}^{o}}\]
B) \[{{45}^{o}}\]
C) \[{{60}^{o}}\]
D) \[{{\tan }^{-1}}\frac{1}{2}\]
Correct Answer: C
Solution :
\[\theta ={{\tan }^{-1}}\left( \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right)={{\tan }^{-1}}\left( \frac{\sqrt{4{{h}^{2}}-4ab}}{a+b} \right)\] \[={{\tan }^{-1}}\left( \frac{\sqrt{3{{a}^{2}}+3{{b}^{2}}+10ab-4ab}}{a+b} \right)=60{}^\circ \].
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