Question

5. If the perimeter of a triangle ABC is 6 times the Arithmetic mean of the sines of its angles. If the side 'a'is 1, then the angle A is?

Answer 1

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°