Higher November 2019 Paper 3 Q18
18 The diagram shows triangle \(ABC\).
\(AB = 3.4\) cm \(AC = 6.2\) cm \(BC = 6.1\) cm
\(D\) is the point on \(BC\) such that
size of angle \(DAC = \dfrac{2}{5} \times\) size of angle \(BCA\)
Calculate the length \(DC\).
Give your answer correct to 3 significant figures.
You must show all your working. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.95 | P1 | for correct substitution into the cosine rule, eg \(3.4^2 = 6.1^2 + 6.2^2 - 2 \times 6.1 \times 6.2 \times \cos BCA\) |
| P1 | for a full process to find \(BCA\) eg \((\cos BCA =)\ \dfrac{6.1^2 + 6.2^2 - 3.4^2}{2 \times 6.1 \times 6.2}\) or \((BCA =)\ 32(.08046913\ldots)\) | |
| P1 | correct substitution into the sine rule, eg \(\dfrac{DC}{\sin\left(\text{``}32.08\ldots\text{''} \times \frac{2}{5}\right)} = \dfrac{6.2}{\sin\left(180 - \text{``}32.08\ldots\text{''} - \left(\text{``}32.08\ldots\text{''} \times \frac{2}{5}\right)\right)}\) | |
| P1 | for complete process to find \(DC\) eg \((DC =)\ \dfrac{6.2 \times \sin \text{``}12.832\text{''}}{\sin \text{``}135.088\text{''}}\) | |
| A1 | Answer in the range 1.94 to 1.951 |
Additional guidance
Can be any angle within triangle ABC
P2 can be awarded for BCA = 32(.08046913…)
Must not come from incorrect processing