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Question

In a triangle ABC, angle A is greater than angle B. If the measures of angles A and B satisfy the equation 3sinx4sin3xk=0,0<k<1, then the measure of angle C is

A
π4
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B
π2
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C
2π3
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D
5π6
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Solution

The correct option is C 2π3
3sinx4sin3x=ksin3x=k

As A and B satisfy this, we get

sin3A=k and sin3B=k

sin3A=sin3B

sin3Asin3B=0

2cos(3A+3B2)sin(3A3B2)=0

Using A+B+C=πA+B=πC

cos(3(πC2))sin(3A3B2)=0

Now as A>Bsin(3A3B2)>0

sin(3C2)=03C2=nπC=2nπ3

For n=1

C=2π3

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