Clock hands

Calculate the internal angles of a triangle whose vertices lie on the clock's 2, 6 and 11 hours.

Correct result:

A =  75 °
B =  60 °
C =  45 °

Solution:

φ=360/12=30 ∘ α=(6−2)⋅ φ=(6−2)⋅ 30=120 ∘ β=(11−6)⋅ φ=(11−6)⋅ 30=150 ∘ γ=(2+12−11)⋅ φ=(2+12−11)⋅ 30=90 ∘  A=β/2=150/2=75∘φ=360/12=30 \ ^\circ \ \\ α=(6-2) \cdot \ φ=(6-2) \cdot \ 30=120 \ ^\circ \ \\ β=(11-6) \cdot \ φ=(11-6) \cdot \ 30=150 \ ^\circ \ \\ γ=(2+12-11) \cdot \ φ=(2+12-11) \cdot \ 30=90 \ ^\circ \ \\ \ \\ A=β/2=150/2=75 ^\circ
B=α/2=120/2=60∘B=α/2=120/2=60 ^\circ
C=γ/2=90/2=45∘C=γ/2=90/2=45 ^\circ

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