Trapezoid height angle

In the isosceles trapezoid ABCD, its bases AB = 20 cm, CD = 12 cm and arms AD = BC = 8 cm are given. Specify its height and alpha angle at vertex A

Final Answer:

v =  6.9 cm
α =  60 °

Step-by-step explanation:

$$\begin{gathered} a=20\text{cm} \\ c=12\text{cm} \\ r=8\text{cm} \\ x=(a-c)/2=(20-12)/2=4\text{cm} \\ v=\sqrt{r^2-x^2}=\sqrt{8^2-4^2}=4\sqrt{3}=6.9\text{cm} \end{gathered}$$

$$\begin{gathered} \sin \alpha =v:r \\ \alpha =\frac{180°}{\pi }\cdot \arcsin (v/r)=\frac{180°}{\pi }\cdot \arcsin (6.9282/8)=60^{\circ } \end{gathered}$$

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