Higher June 2018 Unit 2 Q9
9. ABC and CDE are two right-angled triangles.
In triangle ABC, AB = 6·5 cm and BC = 10·4 cm.
In triangle CDE, CE = 9·4 cm.
\(B\hat{C}E\) = 22°.
\(A\hat{C}B = x^\circ\).
Diagram not drawn to scale
(a) Calculate the value of \(x\). [3]
(b) Hence find the length of DE. [3]
| Scheme | Marks | Comments |
|---|---|---|
| Tan ACB = \(\dfrac{6{\cdotp}5}{10{\cdotp}4}\) | M1 | M1 for equivalent complete method. |
| (ACB =) tan−1 0·625 or tan−1 (6·5 / 10·4) | A1 | |
| (x) = 32(°) | A1 | C.A.O. (Implies previous A1.) Accept an answer that rounds to 32(°) |
Alternative method.
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of ‘two-step’ method. | M2 | A partial trigonometric method is M0. |
| (x) = 32(°) | A1 | Accept an answer that rounds to 32(°) |
| Scheme | Marks | Comments |
|---|---|---|
| (DE =) 9·4 × sin[22 + 32](°) | M2 | FT 22° + ‘their 32°’. M0 for using sin22° or sin ‘their 32°’ alone. M1 for \(\dfrac{\text{DE}}{9{\cdotp}4}\) = sin54(°) |
| = 7·6(…)(cm) ISW | A1 | If no marks awarded SC1 for a correct answer (1dp) using their clearly stated or shown angle (D)C(E), but not 22° or ‘their 32°’. |
Alternative method.
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of ‘two-step’ method. | M2 | A partial trigonometric method is M0. |
| (DE) = 7·6(…)(cm) ISW | A1 |