Question

in the fig o is the center of the circle PQ is the tangent to the circle at A if angle PAB =58 degree find angle ABQ and AQB

Answer 1

Answer:

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Answer 2

Given:

O is the center of the circle

PQ is the tangent to the circle

∠PAB = 58°

To Find:

∠ABQ and ∠AQB

Solution:

∠BAP = ∠APB [alternate opposite angle]

∠BAR = 90°

In ΔABR,

∠ABR + ∠BAR + ∠BRA = 180° [angle sum property of a triangle]

∠ABR = 180 - (58+90)

          = 180- 148

          = 32°

∠ABR = 32°

In ΔABQ,

∠ABR+ ∠BAQ+ ∠AQB = 180° [angle sum property of the triangle]

32 + 122+ ∠AQB = 180°

∠AQB = 180- 154

           = 26°

∠AQB = 26°

Therefore, ∠ABQ = 32° and ∠AQB = 26°.