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Question

If in a ΔABC, (sin A + sin B + sin C)(sin A + sin B – sin C)= 3 sin A sin B, then –

  1. A = 60°    
  2. B = 60°    
  3. C = 60°    
  4. None of these    

hintHint:

use the sine rule to replace the sin angle terms with side lengths and simplify .

The correct answer is: C = 60°


    c=60 degree
    we know that
    a/sin A =b/sin B = c/sinC = 2R
    (sine rule)
    Therefore, we can replace sinA, sin B and sinC as follows:
    (a/2R+b/2R+c/2R)( a/2R+b/2R-c/2R)=3x a/2Rx b/2R
    =>  (a+b+c)(a+b+c)=3ab
    Or
    a2+b2+2ab -c2=3ab
    a2+b2+-c2=ab
    we know that
    cos C = (a2+b2-c2)/2ab
    =ab/2ab = ½
    C = cos-1(1/2)= 60 degree

    the sine rule states that
    a/sin A =b/sin B = c/sinC = 2R
    this is used to find the relation of the angles and sides of the triangles.

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