Higher November 2021 Paper 3 Q29
29 \(A\), \(B\) and \(C\) are three points on the circumference of a circle, centre \(O\).
\(BD\) and \(CD\) are tangents to the circle.
\(ABDC\) is a kite.
Angle \(BDC\) is \(x\)
Not drawn accurately
Prove that angle \(ABO\) is \(\quad 45^\circ - \dfrac{x}{4}\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(OBD\) and \(OCD\) are right angles and \(BOC\) (obtuse) \(= 180 - x\) | M1 | may be on diagram |
| \(BAC = 90 - \dfrac{x}{2}\) | M1dep | oe may be on diagram |
| \(BOC\) (reflex) \(= 180 + x\) and \(ABO + ACO = 360 - \left(90 - \dfrac{x}{2} + 180 + x\right)\) or \(90 - \dfrac{x}{2}\) and \(ABO = \dfrac{1}{2}\left(90 - \dfrac{x}{2}\right)\) \(= 45 - \dfrac{x}{4}\) with M2 scored | A1 | oe \(360 - 90 + \dfrac{x}{2} - 180 - x\) |
| All reasons given tangent meets the radius at 90° angles in a quadrilateral add up to 360° angle at the circumference is half the angle at the centre angles around a point add to 360° | A1 | |
| Alternative method 2 | ||
| \(OBD\) and \(OCD\) are right angles and \(BOC\) (obtuse) \(= 180 - x\) | M1 | may be on diagram |
| \(BAC = 90 - \dfrac{x}{2}\) | M1dep | oe may be on diagram |
| \(BOC\) (reflex) \(= 180 + x\) and \(BAD = \dfrac{1}{2}\left(90 - \dfrac{x}{2}\right)\) or \(45 - \dfrac{x}{4}\) and \(ABO = 180 - \left(45 - \dfrac{x}{4}\right) - \left(90 + \dfrac{x}{2}\right)\) \(= 45 - \dfrac{x}{4}\) with M2 scored | A1 | |
| All reasons given tangent meets the radius at 90° angles in a quadrilateral add up to 360° angle at the circumference is half the angle at the centre angles in a triangle add up to 180° | A1 | |
| Alternative method 3 | ||
| \(OBD\) and \(OCD\) are right angles and \(BOC\) (obtuse) \(= 180 - x\) | M1 | may be on diagram |
| \(BAC = 90 - \dfrac{x}{2}\) | M1dep | oe may be on diagram |
| \(ABC = \dfrac{1}{2}\left[180 - \left(90 - \dfrac{x}{2}\right)\right]\) \(= 45 + \dfrac{x}{4}\) and \(OBC = \dfrac{1}{2}[180 - (180 - x)]\) \(= \dfrac{x}{2}\) and \(ABO = 45 + \dfrac{x}{4} - \dfrac{x}{2}\) \(= 45 - \dfrac{x}{4}\) with M2 scored | A1 | |
| All reasons given tangent meets the radius at 90° angles in a quadrilateral add up to 360° angle at the circumference is half the angle at the centre angles in a triangle add up to 180° (base angles in an) isosceles triangle (are equal) | A1 | |
| Alternative method 4 | ||
| \(OBD\) is a right angle and \(BDO = \dfrac{x}{2}\) | M1 | may be on diagram |
| \(BOD = 90 - \dfrac{x}{2}\) | M1dep | may be on diagram |
| \(OAB + ABO = 90 - \dfrac{x}{2}\) and \(ABO = 45 - \dfrac{x}{4}\) with M2 scored | A1 | |
| All reasons given tangent meets the radius at 90° the diagram is symmetrical oe angles in a triangle add up to 180° exterior angle of a triangle is equal to the sum of the opposite interior angles \(OA\) and \(OB\) are radii, so triangle \(ABO\) is isosceles (base angles in an) isosceles triangle (are equal) | A1 | |
Additional guidance
| Using a value for \(x\) | M0M0A0A0 |