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Question

In a ΔABC, 2sinA+3cosB=32 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=32 ...(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=30B=30A
From eqn(1)
2sinA+3cos(30A)=32
7sinA+33cosA=62 ...(3)

From eqn(2)
3sin(30A)+2cosA=1
7cosA33sinA=2 ...(4)

From eqn(3) and (4), we have
cosA=7+96380.8<32
cosA<cos30
A>30
A+B30
A+B=150
C=30





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