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PA and PB are two tangents to a circle w...

PA and PB are two tangents to a circle with centre 0. from a point P outside the circle. A and B are points on the circle. If `angleAPB= 70^@` then `angleOAB` is equal to:

A

`50^@`

B

`35^@`

C

`20^@`

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 \( \angle APB = 70^\circ \). ### Step-by-Step Solution: 1. **Understanding the Setup**: - We have a circle with center \( O \). - Points \( A \) and \( B \) are points on the circle. - \( PA \) and \( PB \) are tangents to the circle from point \( P \). - The angle \( \angle APB \) is given as \( 70^\circ \). 2. **Using the Properties of Tangents**: - The tangents from a point outside the circle to the circle are equal in length. Therefore, \( PA = PB \). - This implies that triangle \( PAB \) is isosceles with \( PA = PB \). 3. **Finding \( \angle OAP \) and \( \angle OBP \)**: - Since \( PA \) and \( PB \) are tangents to the circle, the angles \( \angle OAP \) and \( \angle OBP \) are right angles (90 degrees). - Thus, we have: \[ \angle OAP = 90^\circ \quad \text{and} \quad \angle OBP = 90^\circ \] 4. **Finding \( \angle APB \)**: - In triangle \( APB \), we know: \[ \angle APB + \angle OAP + \angle OBP = 180^\circ \] - Substituting the known angles: \[ 70^\circ + 90^\circ + 90^\circ = 180^\circ \] - This confirms that the angles are consistent. 5. **Finding \( \angle OAB \)**: - Now, consider triangle \( OAB \). Since \( OA \) and \( OB \) are both radii of the circle, triangle \( OAB \) is also isosceles. - Let \( \angle OAB = \angle OBA = x \). - The sum of angles in triangle \( OAB \) gives: \[ \angle OAB + \angle OBA + \angle AOB = 180^\circ \] - We know \( \angle AOB = 180^\circ - \angle APB = 180^\circ - 70^\circ = 110^\circ \). - Thus, we have: \[ x + x + 110^\circ = 180^\circ \] \[ 2x + 110^\circ = 180^\circ \] \[ 2x = 180^\circ - 110^\circ = 70^\circ \] \[ x = \frac{70^\circ}{2} = 35^\circ \] 6. **Conclusion**: - Therefore, \( \angle OAB = 35^\circ \). ### Final Answer: \[ \angle OAB = 35^\circ \]
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