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PA and PB are tangents to a circle with ...

PA and PB are tangents to a circle with centre O, from a point P outside the circle and A and B are points on the circle. If `/_ APB = 30^(@)`, then `/_ OAB` is equal to `:`

A

A) `40^(@)`

B

B) `15^(@)`

C

C) `50^(@)`

D

D) `25^(@)`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the angle \( \angle OAB \) given that \( \angle APB = 30^\circ \). ### Step-by-Step Solution: 1. **Identify the Given Information**: - We have a circle with center \( O \). - \( PA \) and \( PB \) are tangents from point \( P \) to points \( A \) and \( B \) on the circle. - The angle \( \angle APB = 30^\circ \). 2. **Understanding the Properties of Tangents**: - The tangents from a point outside the circle to the circle are equal in length. Therefore, \( PA = PB \). - The angles formed between the tangent and the radius at the point of tangency are \( 90^\circ \). Thus, \( \angle OAP = \angle OBP = 90^\circ \). 3. **Applying the Triangle Angle Sum Property**: - In triangle \( APB \), we know that the sum of angles in a triangle is \( 180^\circ \). - Let \( \angle OAB = x \). Since \( PA = PB \), we have \( \angle OAP = \angle OBP = 90^\circ \). - Therefore, we can express the angles in triangle \( APB \) as: \[ \angle APB + \angle OAP + \angle OBP = 180^\circ \] \[ 30^\circ + 90^\circ + 90^\circ = 180^\circ \] 4. **Setting Up the Equation**: - Since \( \angle OAP = \angle OBP = 90^\circ \), we can also express \( \angle OAB \) as: \[ \angle OAB + \angle OBA + \angle APB = 180^\circ \] - Let \( \angle OAB = x \) and \( \angle OBA = x \) (since \( OA = OB \)). - Thus, we have: \[ x + x + 30^\circ = 180^\circ \] \[ 2x + 30^\circ = 180^\circ \] 5. **Solving for \( x \)**: - Rearranging the equation gives: \[ 2x = 180^\circ - 30^\circ \] \[ 2x = 150^\circ \] \[ x = 75^\circ \] 6. **Conclusion**: - Therefore, \( \angle OAB = 75^\circ \). ### Final Answer: \[ \angle OAB = 75^\circ \]
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