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 answer:

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

Step-by-step explanation:

$$\begin{gathered} a=73.6 \\ b=57 \\ c=60 \\ d=58.6 \\ {a}_2=a-c=73.6-60={\displaystyle \frac{68}{5}}=13.6 \\ {A}_1=\arccos ((d^2+a_2^2-b^2)\mathrm{/}(2\cdot d\cdot a_2))=\arccos ((58.6^2+13.6^2-57^2)\mathrm{/}(2\cdot 58.6\cdot 13.6))\doteq 1.3366 \\ A=A_1{\to }^{\circ }=A_1\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=1.3366\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=76.58^{\circ }=76.5803^{\circ }=76^{\circ }34^{\mathrm{\prime }}49\mathrm{"} \end{gathered}$$

$$\begin{gathered} B_1=\arccos ((b^2+a_2^2-d^2)\mathrm{/}(2\cdot b\cdot a_2))=\arccos ((57^2+13.6^2-58.6^2)\mathrm{/}(2\cdot 57\cdot 13.6))\doteq 1.5708 \\ B=B_1{\to }^{\circ }=B_1\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=1.5708\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=90^{\circ }=90^{\circ } \end{gathered}$$

$C=180-B=180-90=90^{\circ }=90^{\circ }$

$$\begin{gathered} D=180-A=180-76.5803\doteq 103.4197=103.4197^{\circ }=103^{\circ }25^{\mathrm{\prime }}11\mathrm{"} \\ x=A+B+C+D=76.5803+90+90+103.4197=360 \end{gathered}$$

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