In a trapezium ABCD, as shown, AB//CD, AD=DC=BC=20cm and angle A = 60°, Find all three angles
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Answer: From right triangle ADP
\begin{aligned} \cos 60^{\circ} &=\frac{A P}{A D} \\ \frac{1}{2} &=\frac{A P}{20} \\ A P &=10 \end{aligned}
Similarly from the right triangle BMC, we get BM = 10 cm.
Now from the rectangle PMCD, we get CD = PM = 20 cm.
AB = AP + PM + MB = 10 + 20 + 10 = 40 cm.
(ii)
From the right triangle APD we have
\begin{aligned} \sin 60^{\circ} &=\frac{P D}{20} \\ \frac{\sqrt{3}}{2} &=\frac{P D}{20} \\ P D &=10 \sqrt{3} \end{aligned}
Therefore the distance between AB and CD is 10√3.
Step-by-step explanation:
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