Quadrilaterals,Class-8
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the answer is X=120°. Here you go
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Answer:
∠AOC = 360° - 230° (Complete angle)
∠AOC = 130°
Given that AO = OD = OC,
=> ∠AOD = ∠DOC = ∠AOC/2
=> ∠AOD = ∠DOC = 130°/2 = 65°
Given that AO = OD = OC,
=> ∠ODA = ∠OAD = ∠ODC = ∠OCD = y
(Angles opposite to equal sides are equal)
Consider ΔDOA,
=> ∠ODA + ∠OAD + ∠AOD = 180° (Angle Sum Property
=> y + y + 65° = 180°
=> 2y = 180° - 65°
=> y = 115°/2
Similarly, y = 115°/2 in ΔCOD,
∠ADC = ∠ODA + ∠ODC
=> ∠ADC = y + y
=> ∠ADC = 115°/2 + 115°/2
=> ∠ADC = 230°/2
=> ∠ADC = 115°
AB || DC and AD || BC (from the figure)
=> ABCD is a parallelogram
∠ADC = ∠ABC = x = 115°
(Opposite angles of a parallelogram are equal)
Therefore, x = 115°
Please mark this answer as brainliest.
∠AOC = 360° - 230° (Complete angle)
∠AOC = 130°
Given that AO = OD = OC,
=> ∠AOD = ∠DOC = ∠AOC/2
=> ∠AOD = ∠DOC = 130°/2 = 65°
Given that AO = OD = OC,
=> ∠ODA = ∠OAD = ∠ODC = ∠OCD = y
(Angles opposite to equal sides are equal)
Consider ΔDOA,
=> ∠ODA + ∠OAD + ∠AOD = 180° (Angle Sum Property
=> y + y + 65° = 180°
=> 2y = 180° - 65°
=> y = 115°/2
Similarly, y = 115°/2 in ΔCOD,
∠ADC = ∠ODA + ∠ODC
=> ∠ADC = y + y
=> ∠ADC = 115°/2 + 115°/2
=> ∠ADC = 230°/2
=> ∠ADC = 115°
AB || DC and AD || BC (from the figure)
=> ABCD is a parallelogram
∠ADC = ∠ABC = x = 115°
(Opposite angles of a parallelogram are equal)
Therefore, x = 115°
Please mark this answer as brainliest.
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