Coordinate of A=(3,0,0)
Coordinate of B=(0,-0.75,1.9)
Coordinate of C=(0,1.9,1.5)
Vector AB=(0-3)i+(-0.75-0)j+(1.9-0)k=-3i-0.75j+1.9k
Vector AC=(0-3)i+(1.9-0)j+(1.5-0)k=-3i+1.9j+1.5k
Angle between two vectors is given by
arccos((AB.AC)/magnitude(AB) magnitude (AC))
AB.AC=(-3)(-3)+(-0.75)(1.9)+(1.9)(1.5)=10.425
Magnitude of AB=sqrt(3²+0.75²+1.9²)=3.63
Magnitude of AC=sqrt(3²+1.9²+1.5²)=3.85
Angle=arccos(10.425/(3.63*3.85))=41.8°
This is the required answer
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