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Question

Given:
sin(B+C)=32
sin(A+C)=sin B and
sin A=12.

Find sin(A+BC).

A
12
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B
32
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C
1
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D
12
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Solution

The correct option is B 32
Given,
sin(B+C)=32sin(A+C)=sin Bsin A=12

From sin A=12, we get A=30(1)
sin(B+C)=32, as we know that B+C=60(2)
sin(A+C)=sinB
we get A+C=B(3)
From the equations (1), (2) and (3)
C=60B, A=30
A+C=60B+30=B
2×B=90
B=45
B+C=60,C=6045
C=15
A+BC=30+4515=60
sin(A+BC)=sin 60=32

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