Question

In the given figure, O is the centre of circle, PQ is a tangent to a circle at A. If angle BAQ=60° then find angle ABQ and ANGLE APB..

Answer 1

Answer:

∠APB is 30°.

Step-by-step explanation:

Draw QX tangent to a circle at B.

We know that, tangents drawn from some external point are equal.

Therefore, QA = QB

Δ QBA is an isoceles triangle.

∴ ∠QBA = ∠ QAB

∠QBA = 60°                      (∠QAB = 60°)

⇒ ∠ABQ = 60°

Now, join YO and BY.

∴ ∠ABY = 90°   ( Angle in a semicircle)

In ΔABY, BO is a bisector if ∠B.

∴∠ABO = ∠YBO = 30°

Now, in ΔQBA,  ∠BAP is an exterior angle.

∠BAP = ∠ABQ + ∠BQA

          =   60° + 60°

∠BAP = 120°

Now, in ΔBAP,

∠APB + ∠BAP + ∠ABP = 180°

30° + 120° + ∠ABP = 180°

∠ABP = 30°

Answer 2

Answer:

∠ABQ=60° & ∠APB=30°

Step-by-step explanation: Join OA

We know that, line made at the tangent from the center of the circle is perpendicular, ∠OAB=30°

and, OA=OB(radius of circle) ∠OBA=30°,

and. QA=QB (tangent drawn from an external point are equal)

therefore, ∠QBA=∠QAB=60°

Similarly, ∠BQA=60°

Hence, from ΔQBP, ∠APB=30° ,∠ABQ=60°

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