well the above formula is derived.
Drop a perpendicular from A on to BC and name the point D.
now in triangle CDA, cos 60 = CD/AC = 1/2
CD = 3
thus AD^2 = 6^2-3^2 = 27
now in triangle BDA,
AB^2 = AD^2 + BD^2
BD = 7-3 = 4
thus AB^2 = 16 + 27 = 43.Hence.
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