Four sides of trapezoid

In the trapezoid ABCD is |AB| = 73.6 mm; |BC| = 57 mm; |CD| = 60 mm; |AD| = 58.6 mm. Calculate the size of its interior angles.

Final Answer:

A =  76.5803 °
B =  90 °
C =  90 °
D =  103.4197 °

Step-by-step explanation:

a=73.6 b=57 c=60 d=58.6 a2​=a−c=73.6−60=568​=13.6 A1​=arccos((d2+a22​−b2)/(2⋅ d⋅ a2​))=arccos((58.62+13.62−572)/(2⋅ 58.6⋅ 13.6))≐1.3366 A=A1​→  °=A1​⋅ π180​   °=1.3366⋅ π180​   °=76.58  °=76.5803∘=76°34′49"
B1​=arccos((b2+a22​−d2)/(2⋅ b⋅ a2​))=arccos((572+13.62−58.62)/(2⋅ 57⋅ 13.6))≐1.5708 B=B1​→  °=B1​⋅ π180​   °=1.5708⋅ π180​   °=90  °=90∘
C=180−B=180−90=90∘
D=180−A=180−76.5803≐103.4197∘ x=A+B+C+D=76.5803+90+90+103.4197=360

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planimetricsbasic operations and conceptsgoniometry and trigonometryUnits of physical quantitiesGrade of the word problem

 
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