Question

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'.

Answer 1

The measure of ∠BAD and ∠ADB is 20°20' and 34°90' respectively.

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

`2\angle 2=69^(\circ)80'`

`\angle 2=(69^(\circ)80')/(2)=34^(\circ)90'`

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

`\angle 6=90^(\circ)`

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'

Answer 2

∠BAD=20°20'

∠ADB=34°90'

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'