Question:
if in a triangle ABC, (sinA +sinB+ sinC)(sinA +sinB - sinC) =3 sinA sinB then find the measures of angles of ∆
Answer:
We know that, in a triangle
a/sinA = b/sinB = c/sinC = k(say)
=> sinA = a/k, sinB = b/K, sinC = c/k
Given, (sinA +sinB+ sinC)*(sinA +sinB - sinC) =3 sinA sinB
=> (a/k + b/k+ c/k)*(a/k + b/k - c/k) = 3(a/k)*(b/k)
=> {(a + b + c)*(a + b - c)}/k2 = 3ab/k2
=> (a + b + c)*(a + b - c) = 3ab
=> (a + b)2 - c2 = 3ab
=> a2 + b2 + 2ab - c2 = 3ab
=> a2 + b2 - c2 = 3ab - 2ab
=> a2 + b2 - c2 = ab
=> (a2 + b2 - c2 )/ab = 1
=> (a2 + b2 - c2 )/2ab = 1/2
=> cos C = 1/2 {since cos C = (a2 + b2 - c2 )/2ab}
=> cos C = cos 60
=> C = 60
Now, in a triangle,
A + B + C = 180
=> A + B + 60 = 180
=> A + B = 180 - 60
=> A + B = 120
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