The two tangents drawn from an external point P to a circle with centre o are PA and PB
If ZAPB = 70°, what is the value of ZAOB?
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OAPB forms a quadrilateral.
Therefore, sum of all the angles is 360°
and According to theorem,
Tangent drawn from exterior point to a circle is perpendicular to the circle at the point of contact.
thus, anglePAO & anglePBO =90°
therefore,
anglePAO + anglePBO + angleAPB + angleAOB = 360°
90° + 90° + 70° + angleAOB = 360°
angle AOB = 360° - 250°
angle AOB = 110°
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Therefore, sum of all the angles is 360°
and According to theorem,
Tangent drawn from exterior point to a circle is perpendicular to the circle at the point of contact.
thus, anglePAO & anglePBO =90°
therefore,
anglePAO + anglePBO + angleAPB + angleAOB = 360°
90° + 90° + 70° + angleAOB = 360°
angle AOB = 360° - 250°
angle AOB = 110°
Hope it will help you
Mark it as brainlist
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