Higher January 2022 Paper 2 Q23
23 The diagram shows triangle \(PQR\)
Diagram NOT accurately drawn
\(PQ = 1.6\) cm \(PR = 4.2\) cm Angle \(PRQ = 18°\)
Given that angle \(PQR\) is obtuse,
work out the area of triangle \(PQR\)
Give your answer correct to 3 significant figures.
(6)
| Scheme | Marks |
|---|---|
\(\dfrac{\sin Q}{4.2} = \dfrac{\sin 18}{1.6}\) oe or \(1.6^2 = 4.2^2 + RQ^2 - 2 \times 4.2 \times RQ \times \cos 18\) oe | M1 |
\(\sin^{-1}\left(4.2 \times \dfrac{\sin 18}{1.6}\right)\) (= 54.2) or \(\sin^{-1}(0.811\ldots)\) \(\dfrac{2 \times 4.2 \times \cos 18 \pm \sqrt{(2 \times 4.2 \times \cos 18)^2 - 4 \times 1 \times 15.08}}{2}\) | M1 |
| 180 – “54.2” (=125.8) or \(RQ = 3.0585\ldots\) and \(4.933\ldots\) | M1 |
\((P =)\;180 - \text{``}{125.8}\text{''} - 18\;(= 36.2)\) or \(RQ = \sqrt{4.2^2 + 1.6^2 - 2 \times 4.2 \times 1.6 \times \cos \text{``}{36.2}\text{''}}\;(= 3.0585\ldots)\) or 3.0585 chosen as value from cosine rule above or perpendicular height \(= 4.2\sin \text{``}{36.2}\text{''}\;(= 2.4805\ldots)\) (where “36.2” comes from correct working) | M1 |
\((\text{Area} =)\;\dfrac{1}{2} \times 4.2 \times 1.6 \times \sin(\text{``}{36.2}\text{''})\) or \((\text{Area} =)\;\dfrac{1}{2} \times 4.2 \times \text{``}{3.0585\ldots}\text{''} \times \sin 18\) or \((\text{Area} =)\;\dfrac{1}{2} \times 1.6 \times \text{``}{2.4805\ldots}\text{''}\) | M1 |
| 1.98 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: correct sine ratio - could be rearranged or correct substitution into the cosine rule using angle \(R\)
M1: This can be implied by the correct value(s) (125.8 or 3.0585…) used later
A1: awrt 1.98