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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