A trapezoid

A trapezoid with a base length of a = 36.6 cm, with angles α = 60°, β = 48° and the height of the trapezoid is 20 cm. Calculate the lengths of the other sides of the trapezoid.

Correct result:

b =  23.094 cm
d =  26.9127 cm
c =  5.1356 cm

Solution:

$$\begin{gathered} a=36.6\text{ cm } \\ \alpha =60^{\circ } \\ \beta =48^{\circ } \\ h=20\text{ cm } \\ \sin \alpha =h:b \\ b=h\mathrm{/}\sin (\alpha )=20\mathrm{/}\sin (60^{\circ })=23.094\text{ cm} \end{gathered}$$

$$\begin{gathered} \sin \beta =h:d \\ d=h\mathrm{/}\sin (\beta )=20\mathrm{/}\sin (48^{\circ })=26.9127\text{ cm} \end{gathered}$$

$$\begin{gathered} x_1+c+x_2=a \\ \cos \alpha =x_1:d=x_1:26.9127=0 \\ \cos \beta =x_2:d=x_2:26.9127=0 \\ c=a-d\cdot (\cos {\alpha }^{\circ }+\cos {\beta }^{\circ })=a-d\cdot (\cos 60^{\circ }+\cos 48^{\circ })=36.6-26.9127\cdot (\cos 60^{\circ }+\cos 48^{\circ })=36.6-26.9127\cdot (0.5+0.669131)=5.136=5.1356\text{ cm} \end{gathered}$$

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