A) \[{{90}^{o}}\]
B) \[{{30}^{o}}\]
C) \[{{60}^{o}}\]
D) None of these
Correct Answer: B
Solution :
\[\theta ={{\tan }^{-1}}\left( \frac{-\cot {{30}^{o}}+\cot {{60}^{o}}}{1+\cot {{30}^{o}}\cot {{60}^{o}}} \right)\] \[={{\tan }^{-1}}\left( \frac{\tan {{60}^{o}}-\tan {{30}^{o}}}{1+\tan {{30}^{o}}\tan {{60}^{o}}} \right)={{30}^{o}}\].You need to login to perform this action.
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