Clock hands

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

Result

A =  75 °
B =  60 °
C =  45 °

Solution:

φ=360/12=30  α=(62) φ=(62) 30=120  β=(116) φ=(116) 30=150  γ=(2+1211) φ=(2+1211) 30=90   A=β/2=150/2=75=75φ=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=75 ^\circ
B=α/2=120/2=60=60B=α/2=120/2=60=60 ^\circ
C=γ/2=90/2=45=45C=γ/2=90/2=45=45 ^\circ



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