PA and PB are tangents drawn from a point p to the circle with center O if angle APB equal to 60 than what is angle AOB
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We know that the radius and tangent are perpendicular at their point of contact.
∴ ∠OBP=∠OAP=90
o
Now, In a quadrilateral AOBP
⇒ ∠AOB+∠OBP+∠APB+∠OAP=360
o
[ Sum of four angles of a quadrilateral is 360
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. ]
⇒ ∠AOB+90
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+60
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+90
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=360
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⇒ 240
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+∠AOB=360
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⇒ ∠AOB=120
o
.
Since OA and OB are the radius of a circle then, △AOB is an isosceles triangle.
⇒ ∠AOB+∠OAB+∠OBA=180
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⇒ 120
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+2∠OAB=180
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[ Since, ∠OAB=∠OBA ]
⇒ 2∠OAB=60
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∴ ∠OAB=30
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Step-by-step explanation:
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