In trapezoid 3

In a trapezoid ABCD, the elements are given - lengths of bases a= 20cm, c= 11 cm, angle α = 63°36’ and angle β=79°36’. Calculate the lengths of the other sides and the sizes of the angles.

Final Answer:

b =  13.4576 cm
d =  14.7776 cm
γ =  100.4 °
δ =  116.4 °

Step-by-step explanation:

$$\begin{gathered} a=20\text{cm} \\ c=11\text{cm} \\ \alpha =63°36^{\prime }=63°+\frac{36}{60}°=63.6°=63.6 \\ \beta =79°36^{\prime }=79°+\frac{36}{60}°=79.6°=79.6 \\ \tan \alpha =h:x_1 \\ \tan \beta =h:x_2 \\ {t}_1=\tan \alpha =\tan 63.6°=2.014487=2.01449 \\ {t}_2=\tan \beta =\tan 79.6°=5.448572=5.44857 \\ {x}_1=h/t_1 \\ {x}_2=h/t_2 \\ a=x_1+c+x_2 \\ a=h/t_1+c+h/t_2 \\ h=\frac{a-c}{1/t_1+1/t_2}=\frac{20-11}{1/2.0145+1/5.4486}\doteq 13.2365\text{cm} \\ {x}_1=h/t_1=13.2365/2.0145\doteq 6.5706\text{cm} \\ {x}_2=h/t_2=13.2365/5.4486\doteq 2.4294\text{cm} \\ b=\sqrt{h^2+x_2^2}=\sqrt{13.2365^2+2.4294^2}=13.4576\text{cm} \end{gathered}$$

$d=\sqrt{h^2+x_1^2}=\sqrt{13.2365^2+6.5706^2}=14.7776\text{cm}$

$\gamma =180-\beta =180-79.6=100.4^{\circ }=100°24^{\prime }$

$\delta =180-\alpha =180-63.6=116.4^{\circ }=116°24^{\prime }$

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