Higher November 2018 Unit 2 Q17
17. The diagram below shows two triangles \(ABE\) and \(BCD\), which are constructed from the straight lines \(AD\) and \(EC\) crossing at point \(B\).
\(AE\) = 11 cm, \(EB\) = 13 cm, \(BC\) = 10 cm and \(BD\) = 7 cm.
\(E\hat{A}B\) = 37°.
Diagram not drawn to scale.
Calculate the length \(CD\). [7]
| Scheme | Marks | Comments |
|---|---|---|
| (sin ABE =) \(\dfrac{11 \times \sin 37^\circ}{13}\) | M2 | M1 for \(\dfrac{\sin \text{ABE}}{11} = \dfrac{\sin 37^\circ}{13}\) or equivalent. |
| (ABE =) 30.6(…°) | A1 | Allow 31°. |
| (CBD =) 30.6(…°) | B1 | FT from their derived angle ABE stated or seen on the diagram. Use of 30.6(…°) or ‘their 30.6(…°)’ in subsequent calculation gains B1. |
| (CD =) √[10² + 7² – 2×10×7×cos30.6(…°)] | M2 | M1 for (CD2 = ) 10² + 7² – 2×10×7×cos30.6(…°) FT ‘their CBD’ from either B0 or B1. |
| (CD =) 5.3(...cm) | A1 | Allow an answer of 5.4(cm) from the use of 31°. |