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'
Answer:
The measure of ∠BAD and ∠ADB is 20°20' and 34°90' respectively.
Step-by-step explanation:
Given that 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).
measure of arc BD is 110°20'.
we have to find the measure of ∠BAD and ∠ADB
∠4=110°22'
In ΔOBD, by angle sum property of triangle
∠1+∠2+∠4=180°
∠1+∠2+110°20'=180°
∠1+∠2=69°80'
Since OB=OD(both radii of same circle) therefore ∠1=∠2
[tex]2\angle 2=69^{\circ}80'[/tex]
[tex]\angle 2=\frac{69^{\circ}80'}{2}=34^{\circ}90'[/tex]
m∠ADB=34°90'
As OB is radius of circle and AB is tangent therefore by theorem which states that radius is perpendicular on the tangent line gives
[tex]\angle 6=90^{\circ}[/tex]
By exterior angle property
∠5=∠1+∠2=69°80'
By angle sum property in ΔABO
∠3+∠6+∠5=180°
∠3+90°+69°80'=180°
∠3=20°20'
