In a trapezium ABCD, OC and OD are the bisectors of Angle C and Angle D respectively. Find the measure of Angle ABC and Angle DAB.
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Given OD and OC are bisectors of 2DCB AND ZCDA, respectively.
So, ZOCB = 30°, ZADO = 50° AND AB || CD
In a trapezium,
ZB+C = 180°
→ LB + DCO + LOCB = 180°
→ LB +30° +30° = 180°
→ LB = 180° -60° - 120°
LB = 120°
Similarly, ZA+ZD = 180°
→ ZA+ZODA + 2ODC = 180°
→ ZA +50° +50° = 180°
ZA = 180°- 100° = 80°
Answered by
6
Answer:
Given OD and OC are bisectors of 2DCB AND ZCDA, respectively.
So, ZOCB = 30°, ZADO = 50° AND AB || CD
In a trapezium,
ZB+C = 180°
→ LB + DCO + LOCB = 180°
→ LB +30° +30° = 180°
→ LB = 180° -60° - 120°
LB = 120°
Similarly, ZA+ZD = 180°
→ ZA+ZODA + 2ODC = 180°
→ ZA +50° +50° = 180°
ZA = 180°- 100° = 80°
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