#### Question

In the following figure, PQ is the tangent to the circle at A, DB is the diameter and O is the centre of the circle. If ∠ADB = 30° and ∠CBD = 60°, Calculate:

(i) ∠QAB, (ii) ∠PAD, (iii) ∠CDB,

#### Solution

i) PAQ is a tangent and AB is the chord.

`∠`QAB = `∠`ADB = 30° (angles in the alternate segment)

ii) OA = OD (radii of the same circle)

∴ `∠`OAD =`∠`ODA = 30°

But, OA ⊥ PQ

∴ `∠`PAD = `∠` OAP - `∠` OAD = 90° - 30° = 60°

iii) BD is the diameter.

∴ `∠`BCD = 90 (angle in a semi – circle)

Now in ΔBCD,

`∠` CDB + `∠`CBD + `∠`BCD =180°

⇒ `∠` CDB + 60° + 90° =180°

⇒ `∠`CDB =180° - 150° = 30°

Is there an error in this question or solution?

Solution In the Following Figure, Pq is the Tangent to the Circle at A, Db is the Diameter and O is the Centre of the Circle. If ∠Adb = 30° and ∠Cbd = 60°, Calculate: (I) ∠Qab, (Ii) ∠Pad, (Iii) ∠Cdb, Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.