# 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:

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

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