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Question
Find the measures of all the angles of the parallelogram shown in the figure:
Solution
In ΔBDC,
∠BDC + ∠DCB + ∠CBD = 180°
2a + 5a + 3a = 180°
10a = 180°
⇒ a = 18°
∠BDC = 2a = 2x 18° = 36°
∠DCB = 5a = 5 x 18° = 90°
∠CBD = 3a = 3 x 18° = 54°
∠DAB =∠DCB = 90° ...(opposite angles of parallelogram are equal)
∠DBA = ∠BDC = 36° ...(alternate angles since AB || CD)
∠BDA = ∠CBD = 54° ...(alternate angles since AB || CD)
Therefore, ∠DAB =∠DCB = 90°, ∠DBA + ∠CBD = 90°, ∠BDA + ∠BDC = 90°.
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