A) 0
B) \[90{}^\circ \]
C) \[180{}^\circ \]
D) None of these
Correct Answer: B
Solution :
Here, \[{{\cos }^{2}}\alpha +{{\cos }^{2}}(90-\alpha )+{{\cos }^{2}}\gamma =1\] Þ \[{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha +{{\cos }^{2}}\gamma =1\] Þ \[{{\cos }^{2}}\gamma +1=1\,\,\Rightarrow \,\,\gamma ={{90}^{o}}\].
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