Trapezoid interior angles

Given is an isosceles trapezoid ABCD with bases 10 cm and 14 cm. The height of the trapezoid is 6 cm. Determine the interior angles of the trapezoid.

Final Answer:

α =  71.5651 °
β =  71.5651 °
γ =  108.4349 °
δ =  108.4349 °

Step-by-step explanation:

$$\begin{gathered} a=14\text{cm} \\ c=10\text{cm} \\ h=6\text{cm} \\ x=\frac{a-c}{2}=\frac{14-10}{2}=2\text{cm} \\ \tan \alpha =h:x \\ {\alpha }_1=\arctan (h/x)=\arctan (6/2)\doteq 1.249\text{rad} \\ \alpha ={\alpha }_1\to \text{°}={\alpha }_1\cdot \frac{180}{\pi }\text{°}=1.249\cdot \frac{180}{\pi }\text{°}=71.565\text{°}=71.5651^{\circ }=71°33^{\prime }54" \end{gathered}$$

$\beta =\alpha =71.5651=71.5651^{\circ }=71°33^{\prime }54"$

$\gamma =180-\beta =180-71.5651=108.4349^{\circ }=108°26^{\prime }6"$

$\delta =\gamma =108.4349=108.4349^{\circ }=108°26^{\prime }6"$

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