The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is
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GIVEN: Perimeter of triangle ABC
= 6( sinA + sinB + sinC ) /3 …………(1)
If side opposite to <A, < B, < C are a, b & c respectively.
Then , a+ b+ c = 2( sinA+sinB+ sinC)… by eq (1)
=> 1+b+c= 2( sinA+ sinB+ sinC) ………..(2)
Now, by Sin rule…
a/sinA = b/sinB= c/sinC = k
=> 1/sinA = b/sinB= c/sinC = k
=> 1 = k sinA …….(3)
b= k sinB …….(4)
c = k sinC ……..(5)
By adding (3), (4)& (5)
we get, 1+b+ c = k( sinA+ sinB + sinC)………(6)
By …(2) & …(6)
2(sinA + sinB + sinC) = k( sinS= sinB + sinC)
=> k = 2
So, by eq …..(3)
1 = k SinA
=> sinA = 1/k = 1/2
=> A = 30°
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