in a circle with centre P AB is a minor arc point R is a point other than A and B on major arc AB if angle APB IS 150 THEN FIND ANGLE ARB
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Given, ∠AOB=110
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In △AOB
OA=OB (Radius of the circle)
Thus, ∠OAB=∠OBA (Isosceles triangle property)
Sum of angles of the triangle = 180
∠AOB+∠OAB+∠OBA=180
110+2∠OBA=180
∠OBA=35
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Since, PQ is a tangent touching the circle at B.
Thus, ∠OBQ=90
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Now, ∠ABQ+∠OBA=90
∠ABQ+35=90
∠ABQ=55
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