Higher June 2018 Unit 2 Q19
19. BC is the tangent to the circle at point E, as shown below.
EC = 8 cm, AC = 11 cm and \(D\hat{C}E\) = 31°.
Diagram not drawn to scale
From the formula list: In any triangle \(ABC\): Sine rule \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\); Cosine rule \(a^2 = b^2 + c^2 - 2bc\cos A\)
(a) Calculate the length of AE. [3]
(b) Calculate the size of \(C\hat{E}D\). [4]
| Scheme | Marks | Comments |
|---|---|---|
| (AE2 =) 8² + 11² – 2×8×11×cos31° | M1 | |
| (AE =) 5.8(...cm) | A2 | Award A2 for an answer of 6(cm) from correct working. A1 for (AE2 =) 34.1(…) |
| Scheme | Marks | Comments |
|---|---|---|
| sin CAE= \(\dfrac{8 \times \sin 31^\circ}{5.8(...)}\) | M2 | FT ‘their AE’ from 19(a). Check the diagram. M1 for \(\dfrac{\sin \text{CAE}}{8} = \dfrac{\sin 31^\circ}{5.8(\ldots)}\) or equivalent |
| (CAE=) 44.8(…°) | A1 | Accept answers in the range 44.7° to 45.3°. |
| (CED=) 44.8(…°) | B1 | Strict FT of ‘their CAE’, provided not 31°. Must be convincing (check the diagram). Accept answers in the range 44.7° to 45.3°. |
Alternative method 1
| Scheme | Marks | Comments |
|---|---|---|
| cos(CAE)= \(\dfrac{11^2 + 5.8(\ldots)^2 - 8^2}{2 \times 11 \times 5.8(\ldots)}\) | M2 | FT ‘their AE’ from 19(a). Check the diagram. M1 for \(8^2 = 11^2 + 5.8(\ldots)^2 - 2 \times 11 \times 5.8(\ldots)\ \times\)cos(CAE) |
| (CAE=) 44.8(…°) | A1 | Accept answers in the range 44.7° to 45.3°. |
| (CED=) 44.8(…°) | B1 | Strict FT of ‘their CAE’, provided not 31°. Must be convincing (check the diagram). Accept answers in the range 44.7° to 45.3°. |
Alternative method 2 (Initially evaluating CEA)
| Scheme | Marks | Comments |
|---|---|---|
| sin CEA= \(\dfrac{11 \times \sin 31^\circ}{5.8(...)}\) OR cos CEA= \(\dfrac{5.8(\ldots)^2 + 8^2 - 11^2}{2 \times 5.8(...) \times 8}\) | M2 | FT ‘their AE’ from 19(a). Check the diagram. M1 for \(\dfrac{\sin \text{CEA}}{11} = \dfrac{\sin 31^\circ}{5.8(\ldots)}\) or equivalent OR M1 for 112=5.8(…)2 + 82−2×5.8(…)×8×cosCEA |
| (CEA=) 104.1(…) [or 75.8(…) from sine rule] | A1 | Accept answers in the range 103.7 to 104.3 or 75.7 to 77.7 OR 78. |
| (CAE=) 180−31−75.8(…)= 73.2(…) or 180−31−104.1(…)= 44.8(…) (CED=) 44.8(…) or 73.2(…) | B1 | Strict FT of ‘their CAE’, provided not 31°. Must be convincing (check the diagram). Accept answers in the range 44.7° to 45.3°. |
Corrected by Westie's Workshop
- The board's mark scheme printed “11² = 5.8(…)² + 8² − 2×5.8(…)2×8×cosCEA”. Changed to “11² = 5.8(…)² + 8² − 2×5.8(…)×8×cosCEA” because the cosine rule term is 2 × 5.8(…) × 8 × cos CEA; the extra ‘2’ after 5.8(…) was a stray digit.
Original: see the board's mark scheme PDF (link above).