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$\angle CBE = (180-75-60)^{\circ} = 45 ^{\circ}$
$\angle DEB= (180-65-60)^{\circ}= 55 ^{\circ}$
$\angle GHB= (45+55)^{\circ} = 100^{\circ}$ (exterior angle of a triangle)
$\angle HGC = (100-60)^{\circ} = 40^{\circ}$ (exterior angle of a triangle)
 

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.