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If $\sin A = \sin^{2} B and 2 \cos^{2} A = 3 \cos^{2} B$ then

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sinA = 12

cos B =12

sinC=32

sinC=3+12

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detailed solution

Correct option is A

sin⁡A=sin2⁡B and 2cos2⁡A=3cos2⁡B⇒ 2−2sin2⁡A=3−3sin2⁡B⇒ 2sin2⁡A−3sin⁡A+1=0⇒ (2sin⁡A−1)(sin⁡A−1)=0⇒ A=30∘ or A=90∘if A=30∘⇒B=45∘⇒C=105∘if A−90n⇒B=90∘which is not possible

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If sin⁡A=sin2⁡B and 2cos2⁡A=3cos2⁡B then

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If $\sin A = \sin^{2} B and 2 \cos^{2} A = 3 \cos^{2} B$ then