Question

1) The tangent at A and B of a circle at P. If ∠APQ = 56°, Find ∠PAB.​

Answer 1

Answer:

angle PAB=34 is the right answer

Answer 2

Given :- The tangent at A and B of a circle meets at P. If ∠APQ = 56°, where Q is centre of circle . Find ∠PAB. ?

Solution :-

from image ,

→ ∠APQ = 56° (given.)

So,

→ ∠BPQ = 56° (PQ is angle bisector.)

then,

→ ∠APB = 56 * 2 = 112° .

Now,

→ ∠QAP = ∠QBP = 90° ( Tangents is perpendicular to the radius.)

So, in Quadrilateral APBQ,

→ ∠APB + ∠QAP + ∠QBP + ∠AQB = 360° (angle sum property .)

→ 112° + 90° + 90° + ∠AQB = 360°

→ 292° + ∠AQB = 360°

→ ∠AQB = 360° - 292°

→ ∠AQB = 68° .

Now, in ∆AQB ,

→ ∠AQB = 68°

→ QA = QB = radius of circle.

So,

→ ∠QAB = ∠QBA (angle opposite to equal sides are equal.)

then,

→ ∠AQB + ∠QAB + ∠QBA = 180° (angle sum property.)

→ 68° + ∠QAB + ∠QAB = 180°

→ 2∠QAB = 180° - 68°

→ 2∠QAB = 112°

→ ∠QAB = 56° .

Therefore,

→ ∠PAB = ∠PAQ - ∠QAB

→ ∠PAB = 90° - 56°

→ ∠PAB = 34° (Ans.)

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