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secondary 3 | E Maths
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How to do part (b)
Date Posted: 3 years ago
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Hello there you go.
Date Posted:
3 years ago
Student please note that you have to state the reason why BY = BX ,CZ = CX and AX = AY :
'tangents from an external point are equal in length'
'tangents from an external point are equal in length'
① You'll also have to state why ∠OYB and ∠OZB are 90° :
'angle between tangent and radius of a circle is a right angle'
② Using these and the earlier derived result ∠ABC = 90° from a),
∠YOZ = 360° - 3(90°) = 90° (angle sum of quadrilateral)
So all 4 angles of quadrilateral OYBZ are 90°(right angles)
③ But since OY = OZ (radii of the same circle), and BY = BZ, and all 4 angles are 90° , then all sides are equal and OYBZ is a square.
So OY = OZ = BY = BZ
Since is also a radius of the circle, OX = OY = OZ = BY = BZ
'angle between tangent and radius of a circle is a right angle'
② Using these and the earlier derived result ∠ABC = 90° from a),
∠YOZ = 360° - 3(90°) = 90° (angle sum of quadrilateral)
So all 4 angles of quadrilateral OYBZ are 90°(right angles)
③ But since OY = OZ (radii of the same circle), and BY = BZ, and all 4 angles are 90° , then all sides are equal and OYBZ is a square.
So OY = OZ = BY = BZ
Since is also a radius of the circle, OX = OY = OZ = BY = BZ
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