Question

In the figure ,o is the center of circle if angle adc=140degrees find angle

Answer 1

Answer:

ABCD is a cyclic quardrilateral

ZADC+ZABC=180

140+ZABC=180

ZABC=40

ZACB=90 ( angle of semi circular arc)

In triangle ABC

ZBAC+ZABC+ZACB=180 (angle sum property)

ZBAC=50°

Answer 2

ABCD = cyclic quadrilateral

angleB + angle D = 180°(oppo. sides of quad.)

angleB +140=180°

angleB=180-140

angleB=40°

angle ACB = 90 ° ( angle from arc of semicircle )

now,

angle A +B+C = 180°

angle A + 40°+90°=180°

130+A =180

180-130=50°

angle A = 50 °

Step-by-step explanation:

hopely it will help you