A) \[150{}^\circ \]
B) \[120{}^\circ \]
C) \[60{}^\circ \]
D) \[30{}^\circ \]
Correct Answer: D
Solution :
[d]Given, |
\[2\sin \alpha +3\cos \beta =3\sqrt{2}\] ... (i) |
\[3\sin \beta +2\cos \alpha =1\] ... (ii) |
On squaring and adding Eqs. (i) and (ii), |
we get |
\[4+9+12\,(\sin \alpha \cos \beta +\sin \beta \cos \alpha )=18+1\] |
\[\Rightarrow \]\[12\sin \,(\alpha +\beta )=6\] |
\[\Rightarrow \]\[\sin \,(\alpha +\beta )=\frac{1}{2}\] |
\[\Rightarrow \]\[\alpha +\beta =150{}^\circ \] |
\[\Rightarrow \]\[\alpha +\beta \ne 30{}^\circ \] |
\[\therefore \]\[\gamma =180-(\alpha +\beta )\] |
\[\Rightarrow \]\[\gamma =30{}^\circ \] |
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