A) \[60{}^\circ \]
B) \[90{}^\circ \]
C) \[120{}^\circ \]
D) \[150{}^\circ \]
Correct Answer: C
Solution :
Let \[a=\sin \alpha ,b=\cos \alpha ,c=\sqrt{1+\sin \alpha \cos \alpha }\] then \[\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\] \[\Rightarrow \] \[\cos C=\frac{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +1-\sin \alpha \cos \alpha }{2\sin \alpha \cos \alpha }\] \[\Rightarrow \] \[\cos C=-\frac{\sin \alpha \cos \alpha }{2\sin \alpha \cos \alpha }\] \[\Rightarrow \] \[\cos C=-\frac{1}{2}=\cos 120{}^\circ \] \[\Rightarrow \] \[\angle C=120{}^\circ \]
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