Higher June 2019 Unit 2 Q6
6. \(PQR\) is a right-angled triangle.
\(PR\) = 16·7 cm, \(QR\) = 9·6 cm.
Diagram not drawn to scale
Calculate the size of \(Q\hat{P}R\). [3]
| Scheme | Marks | Comments |
|---|---|---|
| sin (QPR) = \(\dfrac{9{\cdotp}6}{16{\cdotp}7}\) | M1 | |
| (QPR =) sin−1 (9.6/16·7) or sin−1 (0·57..) | m1 | Implies M1. |
| = 35·1(°) or 35·09(°) or 35·089(…°) | A1 | Allow any answer that rounds to 35(°) |
6.Alternative method.
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of ‘two-step’ method. | M2 | A partial trigonometric method is M0. |
| (x) = 35·1(°) or 35·09(°) or 35·089(…°) | A1 | Allow any answer that rounds to 35(°) |