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CBSE Class 10 Standard Math 2025 All Sets Solved Paper

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Question : 2 of 20
Marks: +1, -0
In the adjoining figure, PA and PB are tangents to a circle with centre O. The measure of angle APB is
Solution:  
∠AOB=360∘−210∘\angle AOB = 360^{\circ} - 210^{\circ}
=150∘=150^{\circ}
We know that the radius of a circle is perpendicular to the tangent at the point of Contact.
OA⊥PAOB⊥PBOA \perp PA \quad OB \perp PB
So, ∠PAO=∠PBO=90∘\angle PAO = \angle PBO = 90^{\circ}
The sum of the angles in a quadrilateral is 360∘360^{\circ}
∠AOB+∠PAO+∠PBO+∠APB=360∘\angle AOB + \angle PAO + \angle PBO + \angle APB = 360^{\circ}
150∘+90∘+90∘+∠APB=360∘150^{\circ} + 90^{\circ} + 90^{\circ} + \angle APB = 360^{\circ}
330∘+∠APB=360∘330^{\circ} + \angle APB = 360^{\circ}
∠APB=360∘−330∘\angle APB = 360^{\circ} - 330^{\circ}
∠APB=30∘\angle APB = 30^{\circ}
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