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Question

# In a ΔABC, 2sinA+3cosB=3√2 and 3sinB+2cosA=1. Then the value of angle C is

A
30
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B
60
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C
120
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D
150
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Solution

## The correct option is A 30∘2sinA+3cosB=3√2 ...(1) 3sinB+2cosA=1 ...(2) Squaring eqn(1) and (2), then add ⇒4+9+12sin(A+B)=19 ⇒sin(A+B)=12 ⇒A+B=30∘ or 150∘ If A+B=30∘⇒B=30∘−A From eqn(1) 2sinA+3cos(30∘−A)=3√2 ⇒7sinA+3√3cosA=6√2 ...(3) From eqn(2) 3sin(30∘−A)+2cosA=1 ⇒7cosA−3√3sinA=2 ...(4) From eqn(3) and (4), we have cosA=7+9√638≈0.8<√32 ⇒cosA<cos30∘ ⇒A>30∘ ⇒A+B≠30∘ ∴A+B=150∘ ⇒C=30∘

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