Intermediate June 2018 Unit 2 Q15
15. \(PQR\) is a right-angled triangle, as shown below.
\(PQ = 1{\cdot}41\) m and \(PR = 0{\cdot}89\) m.
Diagram not drawn to scale
Calculate the length of \(QR\). [3]
| Scheme | Marks | Comments |
|---|---|---|
| (QR2 = ) 1·412 + 0·892 | M1 | |
| (QR2) = 2·78(02) or (QR) = √2·78(02) | A1 | Allow 2·8 for 2·78. |
| (QR =) 1·66(…)(m) or 1·67 (m) or 1·7(m) OR 166·7(..) cm or 167cm | A1 | FT from M1 for the correctly evaluated square root of 'their 2.78(02)' provided their answer > 1.41 Allow working in centimetres but penalise −1 from any A marks gained if units not shown for final answer e.g. QR2 = 27802 (A1) , QR = 166·74 (A1) then −1 BUT QR = 166·74 cm OR 167 cm is M1A1A1. |
Alternative method.
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of ‘two-step’ trigonometric relationship. | M2 | A partial trigonometric method is M0. |
| (QR =) 1·66(…)(m) or 1·67 (m) or 1·7(m) OR 166·7(..) cm or 167cm | A1 | C.A.O. |