Rectangular trapezoid

In a rectangular trapezoid ABCD with right angles at vertices A and D with sides a = 12cm, b = 13cm, c = 7cm. Find the angles beta and gamma and height v.

Result

v =  12 cm
B =  67.38 °
C =  112.62 °

Solution:

a=12 cm b=13 cm c=7 cm  x=a−c=12−7=5 cm  v=b2−x2=132−52=12 cm=12 cma=12 \ \text{cm} \ \\ b=13 \ \text{cm} \ \\ c=7 \ \text{cm} \ \\ \ \\ x=a-c=12-7=5 \ \text{cm} \ \\ \ \\ v=\sqrt{ b^2 - x^2 }=\sqrt{ 13^2 - 5^2 }=12 \ \text{cm}=12 \ \text{cm}
B=180∘π⋅arcsin⁡(v/b)=180∘π⋅arcsin⁡(12/13)≐67.3801≐67.38∘≐67∘22′49"B=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(v/b)=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(12/13) \doteq 67.3801 \doteq 67.38 ^\circ \doteq 67^\circ 22'49"
C=180−B=180−67.3801=563150=112.62=112.62∘=112∘37′12"C=180-B=180-67.3801=\dfrac{ 5631 }{ 50 }=112.62=112.62 ^\circ =112^\circ 37'12"

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