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

Solution:

a = 73.6 b = 57 c = 60 d = 58.6 a_{2} = a - c = 73.6 - 60 = \frac{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} - 5 7^{2}) / (2 \cdot 58.6 \cdot 13.6)) ≐ 1.3366 A = A_{1} \to^{\circ} = A_{1} \cdot {\frac{180}{π}}^{\circ} = 1.3366 \cdot {\frac{180}{π}}^{\circ} = 76.5 8^{\circ} = 76.580 3^{\circ} = 7 6^{\circ} 3 4^{'} 49 "

B_{1} = \arccos ((b^{2} + a_{2}^{2} - d^{2}) / (2 \cdot b \cdot a_{2})) = \arccos ((5 7^{2} + 13. 6^{2} - 58. 6^{2}) / (2 \cdot 57 \cdot 13.6)) ≐ 1.5708 B = B_{1} \to^{\circ} = B_{1} \cdot {\frac{180}{π}}^{\circ} = 1.5708 \cdot {\frac{180}{π}}^{\circ} = 9 0^{\circ}

C = 180 - B = 180 - 90 = 9 0^{\circ}

D = 180 - A = 180 - 76.5803 ≐ 103.4197 = 103.419 7^{\circ} = 10 3^{\circ} 2 5^{'} 11 " x = A + B + C + D = 76.5803 + 90 + 90 + 103.4197 = 360

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