In an acute-angled triangle ABC the angle at A measures 40°. I is the incentre and O is the circumcentre. By how much does ∠BIC exceed ∠BOC?

Two different rules are at work. The incentre gives ∠BIC = 90° + ∠A/2, because in triangle BIC the angles at B and C are halves of the triangle's own angles: ∠BIC = 180° − (∠B + ∠C)/2 = 180° − (180° − 40°)/2 = 110°. The circumcentre gives the central angle ∠BOC = 2∠A = 80°, twice the inscribed angle on the same chord BC. The difference is 110° − 80° = 30°. The trap 20° is ∠A/2 and 40° is ∠A itself, each got by mixing the two rules into one; 10° halves the correct difference.