Question

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AB is tangent to the circle k(O) at B, and AD is a secant, which goes through O. Point O is between A and D∈k(O). Find m∠BAD and m∠ADB, if measure of arc BD is 110°20'.

Answer 1

Answer:

∠BAD = 20°20'

∠ADB = 34°50'

Step-by-step explanation:

From the figure,

ΔOBD is an isoceles triangle.

⇒ ∠ODB = ∠OBD

Sum of angles a triangle is 180°

⇒ ∠ODB +∠OBD + ∠BOD = 180°

⇒ ∠ODB +∠ODB + ∠BOD = 180°

⇒ 2∠ODB + 110°20' = 180°

⇒ 2∠ODB = 179°60'-110°20'

⇒ 2∠ODB = 69°40'

⇒ 2∠ODB = 68°100'

⇒ ∠ODB = 34°50'

Therefore, ∠ADB = 34°50'

since OB ⊥ AB,

∠OBD + ∠DBB' = 90°

34°50' + ∠DBB' = 90°

∠DBB' = 89°60'-34°50'

∠DBB' = 55°10'

We know that, exterior angle of a triangle is equal to sum of interior opposite angles

Then,

∠DBB' = ∠ADB + ∠BAD

55°10' = 34°50' + ∠BAD

54°70' - 34°50' = ∠BAD

20°20' = ∠BAD

∠BAD = 20°20'