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In Figure, if PA and PB are tangents to the circle with centre O such that \angleAPB = 50°, then \angleOAB is equal to

(A) 25°                        (B) 30°                        (C) 40°                        (D) 50°

Answers (1)

Answer (A) 25°
Given : \angleAPB = 50°
  We know that length of tangents drawn from an external point is equal
Hence, PA = PB
Since, PA = PB
Let  \anglePAB = \angle PBA = x0     
In \bigtriangleupPAB                 
\anglep+\angleA+\angleB = 180^{\circ}     (\because  Sum of interior angles of a tangent is 180^{\circ})
50^{\circ}+x^{\circ}+x^{\circ}= 180^{\circ}
2x^{\circ}= 130^{\circ}
x^{\circ}= 65^{\circ}

\anglePAB = \anglePBA = 65^{\circ}
\anglePAO = 90^{\circ} (\mathbb{Q}  tangent is perpendicular to radius)
\anglePAO = \anglePAB +\angleOAB
90^{\circ}= 65^{\circ}+\angle OAB
\angle OAB= 90^{\circ}-65^{\circ}
\angle OAB= 25^{\circ}


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