answer the 3 Question
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manavi88:
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in quadrilateral P0RQ
Angle POR = 90° + (angleQ / 2)
Angle POR = 90° + (100/ 2)
= 90° + 50° = 140°
Angle POR = 140°
AngleOPR and angle ORP will be equal acute angles.
in triangle POR,
Angle OPR + angleORP + 140° = 180°
angleOPR + Angle ORP = 40°
since , PO = OR ( radii of circle are always equal)
so then ,
AngleOPR = angleORP = 20°
Answer:
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AngleOPR = 20° , angle ORP = 20°
Angle POR = 90° + (angleQ / 2)
Angle POR = 90° + (100/ 2)
= 90° + 50° = 140°
Angle POR = 140°
AngleOPR and angle ORP will be equal acute angles.
in triangle POR,
Angle OPR + angleORP + 140° = 180°
angleOPR + Angle ORP = 40°
since , PO = OR ( radii of circle are always equal)
so then ,
AngleOPR = angleORP = 20°
Answer:
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AngleOPR = 20° , angle ORP = 20°
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