Question 1375

 Even we cannot have good diagram of triangle here, try to figure
 out if angle BAC is 120 deg, both ABC & BCA are 30 deg, and AB=AC = 7.
       A
      /|\
     / | \
    /  |  \
   B   D   C

 Let AD be the height of the base BC(note AD is perendicular to BC),
 we ses that angle BAD = 120/2 =60, so BD/AB = sin 60 , 
 BD = AB sin 60 = 7 * sqrt(3)/2 = 6.06 
 Since BD=CD,BC = 2 BD.
 Hence,the longest leg BC = 2*BD = 12.12 

 Of course, we can use the law of cosine directly, [Note: a=BC, b= AC, c=AB]
 a^2 =b^2 + c^2  - 2bc cos A,
     = 7^2 + 7^2 - 2*7^2 cos(120)
     = 98(1 +0.5) [Note: cos 120 = -0.5]
     = 147,
 So, a = sqrt(147) = 12.12 (same answer)