Question

In the figure, AB is the diameter of the circle with centre O. If ∠DAB = 70° and ∠DBC = 30°, determine ∠ABD, ∠CDB

Answer 1

ABCD IS A CYCLIC quad.
so, <DAB +< DCB = 180
70 + < DCB =180
< DCB = 110
IN TRIANGLE DBC
< DCB + < CDB + <DBC = 180 ( SUM OF ALL ANGLES IN TRIANGLE IS 180)
110 + 30 +<CDB =180
< CDB = 40
<ADC = 90 + < BDC ( ANGLE IN A SEMICIRCLE IS 90)
<ADC + < ABD +<DBC = 180 ( OPPO. ANGLE OF CYCLIC QUAD.)
90 +40 +30 + < ABD = 180
<ABD =20

Answer 2

♠ ANSWER :

REFER THE ATTACHMENT.