Higher November 2020 Paper 2 Q23
23.
Diagram not drawn to scale
In the diagram, \(AD\) = 5·7 cm, \(BD\) = 9·6 cm, \(B\hat{D}C=79^\circ\) and \(D\hat{B}C=39^\circ\).
\(ADC\) is a straight line.
(a) Calculate the length of \(DC\). [3]
(b) Mona assumes that the values in the diagram are all exact and uses these to work out the area of triangle \(ABD\).
In fact, the lengths are correct but \(B\hat{D}C\) has been rounded up to the nearest whole number.
Is Mona’s answer too large or too small?
Use calculations to justify your decision. [3]
In fact, the lengths are correct but \(B\hat{D}C\) has been rounded up to the nearest whole number.
Is Mona’s answer too large or too small?
Use calculations to justify your decision. [3]
Too large ☐ Too small ☐
| Scheme | Marks | Comments |
|---|---|---|
| \(\mathrm{DC}=\dfrac{9.6}{\sin(180-(79+39))}\times\sin39\) | M2 | M1 for \(\dfrac{\mathrm{DC}}{\sin39}=\dfrac{9.6}{\sin(180-(79+39))}\) |
| 6.8(…) (cm) | A1 |
| Scheme | Marks | Comments |
|---|---|---|
| \(A\hat{D}B \gt 101\) | B1 | Accept example of \(101\lt A\hat{D}B\lt101.5\) |
| \(\sin A\hat{D}B \lt \sin101\) | B1 | Accept FT example of \(0.9799\lt\sin A\hat{D}B\lt0.9816\) |
| Mona’s area is \(\tfrac{1}{2}\times9.6\times5.7\sin A\hat{D}B\) and is too large or \(\tfrac{1}{2}\times AD\times BD\times\sin A\hat{D}B\) is too large | B1 | Need both ‘too big’ and sight of \(\tfrac{1}{2}ab\sin C\). Accept calculation using \(\tfrac{1}{2}\times AD\times BD\times\sin A\hat{D}B\) e.g. \(26.810\lt\text{area}\lt26.857\) If no marks award SC1 for a convincing explanation without calculations, e.g. by drawing B3 for Area = \(\tfrac{1}{2}\times9.6\times5.7\sin A\hat{D}B\) and \(\sin101\gt\sin A\hat{D}B\) |
| (6) |