Question

In given figure, AB is a diameter of the semi-circle. Chord DC II AB Given that ABC = 55°, find
(i) Angle CAB
(ii) Angle DEA




Sorry for Ganda diagram.....​

Answer 1

Answer:

1) in triangle CAB

angle ACB = 90° ( diameter subtends 90° on circumference)

angle ABC = 55° ( given)

ABC + CAB + ACB = 180° ( angle sum property)

CAB = 35°

2) as DE || AB , angle CAB = angle ACD ( alternate)

angle ACD = 35°

now in quad AECD ,angle ACD and DEA are supplementary as they are opposite

so Angle DEA = 180° - 35°

= 145°

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