From an External Point P , Tangents Pa = Pb Are Drawn to a Circle with Centre O . If ∠ P a B = 50 O , Then Find ∠ a O B - Mathematics

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Answer in Brief

From an external point P , tangents PA PB are drawn to a circle with centre O   . If  \[\angle PAB = {50}^o\] , then find  \[\angle AOB\]

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Solution

It is given that PA and PB are tangents to the given circle.

\[\therefore \angle PAO = 90^o\]   (Radius is perpendicular to the tangent at the point of contact.)
Now,
\[\angle PAB = 50^o\]          (Given)
\[\therefore \angle OAB = \angle PAO - \angle PAB = 90^o - 50^o = 40^o\]
In ∆OAB,
OB = OA    (Radii of the circle)
\[\therefore \angle OAB = \angle OBA = 40^o\] (Angles opposite to equal sides are equal.)
Now,
\[\angle AOB + \angle OAB + \angle OBA = 180^o\] (Angle sum property) 
\[\Rightarrow \angle AOB = 180^o - 40^o - 40^o= 100^o\]
  Is there an error in this question or solution?
Chapter 8: Circles - Exercise 8.2 [Page 37]

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RD Sharma Class 10 Maths
Chapter 8 Circles
Exercise 8.2 | Q 32 | Page 37

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