Foundation June 2024 Paper 1 Q16
16. In the diagram below, \(BE\) is a straight line.
Diagram not drawn to scale
Show that \(x\) = 150°.
You must give a reason for each step of your working. [4]
| Scheme | Marks | Comments |
|---|---|---|
| \(A\hat{C}B\) = 60(°) or \(A\hat{B}C\) = 60(°) or \(B\hat{A}C\) = 60(°) AND \(A\hat{C}D\) = 120(°) AND \(A\hat{D}C\) = 30(°) | B2 | Angles may be seen on diagram. B1 for any two of the three angles 30(°), 120(°) and 60(°) |
| A correctly labelled and explained solution \(A\hat{C}B\) = 60(°) equilateral triangle \(A\hat{C}D\) = 120(°) angles on a straight line \(A\hat{D}C\) = 30(°) angles in an isosceles triangle (so \(x\) = 150°) OR starting with the 150(°) \(A\hat{D}C\) = 30(°) angles on a straight line \(A\hat{C}D\) = 120(°) angles in an isosceles triangle \(A\hat{C}B\) = 60(°) so \(ABC\) is an equilateral triangle OR working from both sides \(A\hat{C}B\) = 60(°) as an equilateral triangle \(A\hat{D}C\) = 30(°) angles on a straight line \(A\hat{C}D\) = 120(°) angles in an isosceles triangle AND 120 + 60 = 180 angles on a straight line | E2 | E2 and E1 dependent on B2 being awarded. Accept \(A\hat{C}B\) = 60(°) as ‘all angles (in \(ABC\)) are equal/60°’ Accept \(A\hat{D}C\) = 30(°) as ‘two angles (in \(ACD\)) are equal’ E1 for either:
|
| (4) |
Corrected by Westie's Workshop
- The board's mark scheme printed “as all angles (in ABC) are equal/60°’”. Changed to “as ‘all angles (in ABC) are equal/60°’” because the quoted reason had a closing quote mark but no opening one.
- The board's mark scheme printed “as two angles (in ACD) are equal’”. Changed to “as ‘two angles (in ACD) are equal’” because the quoted reason had a closing quote mark but no opening one.
- The board's mark scheme printed “working from both side”. Changed to “working from both sides” because “side” should be “sides”.
Original: see the board's mark scheme PDF (link above).