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.

Correct result:

A =  76.58 °
B =  90 °
C =  90 °
D =  103.42 °

Solution:

a=73.6 b=57 c=60 d=58.6 a2=a−c=73.6−60=685=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⋅ 180π  ∘=1.3365788376612⋅ 180π  ∘=76.58  ∘=76.58∘=76∘34′49"a=73.6 \ \\ b=57 \ \\ c=60 \ \\ d=58.6 \ \\ a_{2}=a-c=73.6-60=\dfrac{ 68 }{ 5 }=13.6 \ \\ A_{1}=\arccos((d^2+a_{2}^2-b^2)/(2 \cdot \ d \cdot \ a_{2}))=\arccos((58.6^2+13.6^2-57^2)/(2 \cdot \ 58.6 \cdot \ 13.6)) \doteq 1.3366 \ \\ A=A_{1} \rightarrow \ ^\circ =A_{1} \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =1.3365788376612 \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =76.58 \ \ ^\circ =76.58 ^\circ =76^\circ 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⋅ 180π  ∘=1.5707963267949⋅ 180π  ∘=90  ∘=90∘B_{1}=\arccos((b^2+a_{2}^2-d^2)/(2 \cdot \ b \cdot \ a_{2}))=\arccos((57^2+13.6^2-58.6^2)/(2 \cdot \ 57 \cdot \ 13.6)) \doteq 1.5708 \ \\ B=B_{1} \rightarrow \ ^\circ =B_{1} \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =1.5707963267949 \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =90 \ \ ^\circ =90 ^\circ
C=180−B=180−90=90∘C=180-B=180-90=90 ^\circ
D=180−A=180−76.5803=517150=103.42=103.42∘=103∘25′12" x=A+B+C+D=76.5803+90+90+103.42=360D=180-A=180-76.5803=\dfrac{ 5171 }{ 50 }=103.42=103.42 ^\circ =103^\circ 25'12" \ \\ x=A+B+C+D=76.5803+90+90+103.42=360

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