Higher November 2021 Unit 2 Q12
12. The diagram shows a quadrilateral DEFG.
Diagram not drawn to scale
Calculate the size of \(E\hat{F}G\). [7]
From the formula list: In any triangle \(ABC\): Sine rule \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\); Cosine rule \(a^2 = b^2 + c^2 - 2bc\cos A\)
| Scheme | Marks | Comments |
|---|---|---|
| Use of cosine rule followed by sine rule | S1 | |
| \((EG =)\) \(\sqrt{(2{\cdot}7^2 + 3{\cdot}2^2 - 2 \times 2{\cdot}7 \times 3{\cdot}2 \times \cos 79(^\circ))}\) | M2 | M1 for \((EG^2 =)\ 2{\cdot}7^2 + 3{\cdot}2^2 - 2 \times 2{\cdot}7 \times 3{\cdot}2 \times \cos 79(^\circ)\) or for \((EG^2 =)\ 14{\cdot}2(3\ldots)\) |
| \((EG =)\ 3{\cdot}77\ldots\) (cm) | A1 | Accept 3·8 cm Allow \(\sqrt{14.2(3\ldots)}\) if used in this form in subsequent work, provided not evaluated as a decimal (at any stage) |
| \(\sin EFG = EG \times \sin 65(^\circ)/6{\cdot}4\) OR \(EFG = \sin^{-1}(EG \times \sin 65(^\circ)/6{\cdot}4)\) | M2 | F.T. ‘their derived \(EG\)’ (not 2·7, 3·2, 6·4 or spurious \(EG\)). Award M1 for \(\sin EFG/EG = \sin 65(^\circ)/6{\cdot}4\) OR \(EG/\sin EFG = 6{\cdot}4/\sin 65(^\circ)\) |
| \(F = 32(\cdot29\ldots^\circ)\) | A1 | Dependent on previous M2. |