If `a, b, A` be given in a triangle and `c_1 and c_2` be two possible value of the third side such that `c_1^2+c_1c_2+c_2^2=a^2,` then a is equal to

If a , b ,\ A are given in a triangle and c_1a n d\ c_2 are two possible values of third side such that c_1^2+c_1c_2+c_2^2=a^2 , then A is equal to 30^0 (b) 60^0 (c) 90^0 (d) 120^0

If a, b and A are given in a triangle and c_(1), c_(2) are possible values of the third side, then prove that c_(1)^(2) + c_(2)^(2) - 2c_(1) c_(2) cos 2A = 4a^(2) cos^(2)A

If a, b and A are given in a triangle and c_(1), c_(2) are possible values of the third side, then prove that c_(1)^(2) + c_(2)^(2) - 2c_(1) c_(2) cos 2A = 4a^(2) cos^(2)A

If a, b and A are given in a triangle and c_(1), c_(2) are possible values of the third side, then prove that c_(1)^(2) + c_(2)^(2) - 2c_(1) c_(2) cos 2A = 4a^(2) cos^(2)A

If a, b and A are given in a triangle and c_(1), c_(2) are possible values of the third side, then prove that c_(1)^(2) + c_(2)^(2) - 2c_(1) c_(2) cos 2A = 4a^(2) cos^(2)A

If a, b and A are given in a triangle and c_(1), c_(2) are possible values of the third side, then prove that c_(1)^(2) + c_(2)^(2) - 2c_(1) c_(2) cos 2A = 4a^(2) cos^(2)A

If a,b and /_A are given in a triangle and c_(1),c_(2) are the possible values of the remaining side,then c_(1)^(2)+c_(2)^(2)-2c_(1)c_(2)cos2A is equal to:

In a triangle a = sqrt5,b = 2, /_A=pi/6 and the two possible values of third side are c_1 and c_2 , then

In the ambiguous case, if a, b and A are given and c_1, c_2 are the two values of the third (c_1-c_2)^2 + (c_1+c_2)^2 tan^2 A is equal to

In an ambiguous case of solving a triangle when a=sqrt5,b=2,angleA=pi/6 and the two possible value of third side are c_1 and c_2 then