Higher June 2019 Unit 1 Q14
14. The points \(A\), \(B\) and \(C\) lie on the circumference of a circle.
The straight lines \(EBD\) and \(ECF\) are tangents to the circle.
\(B\hat{E}C = 58^\circ\) and \(B\hat{C}A = 35^\circ\).
Diagram not drawn to scale
Find the size of \(A\hat{B}C\).
You must show all your working. [4]
| Scheme | Marks | Comments |
|---|---|---|
| Check diagram. | ||
| EBC or ECB = (180 – 58) / 2 | M1 | Angles in an isosceles triangle. |
| = 61(°) | A1 | |
| BAC = 61(°) | B1 | Alternate segment theorem. FT ‘their EBC or ECB’. |
| ABC (= 180 – 35 – 61) = 84(°) | B1 | FT 180 – 35 – ‘their BAC’. |
Alternative method 1
| Scheme | Marks | Comments |
|---|---|---|
| Check diagram. | ||
| EBC or ECB = (180 – 58) / 2 | M1 | Angles in an isosceles triangle. |
| = 61(°) | A1 | |
| DBA = 35(°) | B1 | Alternate segment theorem. |
| ABC (= 180 – 35 – 61) = 84(°) | B1 | Angles on a straight line FT 180 – ‘their EBC’ – ‘their DBA’. |
Alternative method 2
| Scheme | Marks | Comments |
|---|---|---|
| Check diagram. | ||
| EBC or ECB = (180 – 58) / 2 | M1 | Angles in an isosceles triangle. |
| = 61(°) | A1 | |
| ACF (=180 – 35 – 61) = 84(°) | B1 | Angles on a straight line. FT 180 – 35 – ‘their ECB’. |
| ABC = 84(°) | B1 | Alternate segment theorem. FT ‘their ACF’. |
Alternative method 3
| Scheme | Marks | Comments |
|---|---|---|
| (using isosceles triangle BOC, where O is the centre of the circle) | Check diagram. | |
| BOC = 360 – 90 – 90 – 58 | M1 | Angles in kite BOCE |
| = 122 | A1 | |
| BAC = 61 | B1 | Use of angle in the centre FT ‘their BOC’ ÷ 2 |
| ABC (= 180 – 35 – 61) = 84(° | B1 | FT 180 – 35 – ‘their BAC’ |