Could anyone help, please?
1 answer:
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Answer:
∠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'
