Ab is tangent to the circle k(o) at b, and ad is a secant, which goes through center o. point o is between a and d∈k(o). find m∠bad and m∠adb, if the measure of arc bd is 110°20'.

∠BAD=20°20'

∠ADB=34°90'

Step-by-step explanation:

AB is tangent to the circle k(O), then AB⊥BO. If the measure of arc BD is 110°20', then central angle ∠BOD=110°20'.

Consider isosceles triangle BOD (BO=OD=radius of the circle). Angles adjacent to the base BD are equal, so ∠DBO=∠BDO. The sum of all triangle's angles is 180°, thus

∠BOD+∠BDO+∠DBO=180°

∠BDO+∠DBO=180°-110°20'=69°80'

∠BDO=∠DBO=34°90'

So ∠ADB=34°90'

Angles BOD and BOA are supplementary (add up to 180°), so

∠BOA=180°-110°20'=69°80'

In right triangle ABO,

∠ABO+∠BOA+∠OAB=180°

90°+69°80'+∠OAB=180°

∠OAB=180°-90°-69°80'

∠OAB=20°20'

So, ∠BAD=20°20'

is it 35

step-by-step explanation: