A circular metal disc used as a stationery paperweight has centre \(O\). Points \(D\) and \(E\) lie on the circumference, and tangents to the circle at \(D\...
Question 1 Report
A circular metal disc used as a stationery paperweight has centre \(O\). Points \(D\) and \(E\) lie on the circumference, and tangents to the circle at \(D\) and \(E\) meet at an external point \(F\), shown in the diagram. The angle between the tangents, angle \(DFE\), is \(46^{\circ}\).
- Write down the size of angle \(ODF\) and angle \(OEF\), giving a reason. (1)
- Work out the size of angle \(DOE\). (2)
- Hence work out the size of angle \(DEO\), giving a reason. (2)
Answer Details
This question again uses the tangent-radius right angle, then the quadrilateral angle sum, and finally isosceles-triangle reasoning within the resulting figure.
Both angle \(ODF\) and angle \(OEF\) are \(90^\circ\), because a tangent meets a radius at right angles [1 mark].
The four angles of quadrilateral \(ODFE\) sum to \(360^\circ\):
\[\text{angle } DOE = 360^\circ - 90^\circ - 90^\circ - 46^\circ = 134^\circ\][2 marks]
Triangle \(ODE\) is isosceles, since \(OD\) and \(OE\) are both radii of the circle, so the two base angles at \(D\) and \(E\) are equal, sharing the remaining angle:
\[\text{angle } DEO = \dfrac{180^\circ - 134^\circ}{2} = 23^\circ\][2 marks]
Every radius drawn to a point of tangency creates a right angle with that tangent; combined with any two tangents from the same external point being equal in length (making triangle \(ODE\) isosceles), this pair of facts underpins nearly every "two tangents from a point" question.
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