please solve this question 27
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OA AND OB are the radii hence OA = OB
so the triangle AOB is isoceles .
Hence by property of isoceles triangle , Angle BAO = Angle OBA = 45
now , OA is also a tangent at A hence By tangent property , Angle QAO = 90
Hence we have Angle Angle BAQ = 45.
So now we have Angle AOB = 90 by AMS property of triangle.
Therefore we know that the angle formed at center of circle is twice the angle formed elsewhere.
Hence Angle ACB = 90 / 2 = 45
Angle ACB = Angle BAQ = 45
so the triangle AOB is isoceles .
Hence by property of isoceles triangle , Angle BAO = Angle OBA = 45
now , OA is also a tangent at A hence By tangent property , Angle QAO = 90
Hence we have Angle Angle BAQ = 45.
So now we have Angle AOB = 90 by AMS property of triangle.
Therefore we know that the angle formed at center of circle is twice the angle formed elsewhere.
Hence Angle ACB = 90 / 2 = 45
Angle ACB = Angle BAQ = 45
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In triangale, AOB angle O is 90°
then angle B is 45° and let the angle of A be x
then, acc to angel sum property in triangle
90°+45°+x=180°
x=45°
PAQ is a tangent then angel OAQ must be 90°
angel BAQ=45°
by tangent coard therom
Angel BAQ=angle ACB
then angle B is 45° and let the angle of A be x
then, acc to angel sum property in triangle
90°+45°+x=180°
x=45°
PAQ is a tangent then angel OAQ must be 90°
angel BAQ=45°
by tangent coard therom
Angel BAQ=angle ACB
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