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a
30∘
b
60∘
c
120∘
d
150∘
answer is A.
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Detailed Solution
cos2θ=1−sin2θ. Let 81sin2θ=z, we get z+81z=30 or z2−30z+81=0 or (z−27)(z−3)=0 i.e., 81sin2θ=34sin2θ=33 or 31∴ sin2θ=34,14 or sinθ=±32,±12∴ θ=30∘,60∘,120∘,150∘