Higher November 2018 Unit 2 Q9
9. The diagram below shows two right-angled triangles \(ABC\) and \(APQ\).
\(AB\) = 8·2 cm, \(BC\) = 6·4 cm and \(PQ\) = 7·9 cm.
\(C\hat{A}Q\) = 90°.
Diagram not drawn to scale
(a) Calculate the size of angle \(x\). [3]
(b) Calculate the length \(AQ\). [4]
| Scheme | Marks | Comments |
|---|---|---|
| Tan x = \(\dfrac{6{\cdot}4}{8{\cdot}2}\) | M1 | |
| ( x =) tan−1 0·78(0..) or tan-1 \(\dfrac{6{\cdot}4}{8{\cdot}2}\) | A1 | |
| = 38(°) OR 37·9(...°) | A1 | Implies previous A1. |
Alternative method.
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of ‘two-step’ method. | M2 | A partial trigonometric method is M0. |
| (x) = 38(°) | A1 | Accept an answer that rounds to 38(°) |
| Scheme | Marks | Comments |
|---|---|---|
| (PAQ = 90 – 38 =) 52(°) | B1 | FT 90° − ‘their 38°’. May be seen on the diagram. |
| AQ = \(\dfrac{7{\cdot}9}{\sin 52(^\circ)}\) | M2 | FT ‘their clearly defined PAQ’ BUT not if PAQ = ‘their x’. M1 for sin52(°) = \(\dfrac{7{\cdot}9}{\text{AQ}}\) |
| (AQ) = 10(cm) OR 10·0(…cm) | A1 |
Alternative method.
| Scheme | Marks | Comments |
|---|---|---|
| PQA =38(°) | B1 | FT ‘their 38°’. May be seen on the diagram. |
| AQ = \(\dfrac{7{\cdot}9}{\cos 38(^\circ)}\) | M2 | FT ‘their clearly defined PQA’ M1 for cos 38(°) = \(\dfrac{7{\cdot}9}{\text{AQ}}\) |
| (AQ) = 10(cm) OR 10·0(…cm) | A1 |
Alternative method.
| Scheme | Marks | Comments |
|---|---|---|
| (PAQ = 90 – 38 =) 52(°) | B1 | FT 90° − ‘their 38°’. |
| Correct use of ‘two-step’ method. | M2 | A partial trigonometric method is M0. FT ‘their clearly defined PAQ’ BUT not if PAQ = ‘their x’. |
| (AQ) = 10(cm) | A1 | Accept an answer that rounds to 10(cm) |