Intermediate June 2024 Unit 2 Q16
16. In the diagram below, \(ABD\) and \(BCD\) are both right-angled triangles.
Diagram not drawn to scale
\(AB\) = 7·3 cm, \(AD\) = 13·8 cm and \(CD\) = 5·5 cm.
Calculate the value of \(y\). [6]
| Scheme | Marks | Comments | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| 13·82 = \(BD\)2 + 7·32 OR (\(BD\)2 =) 13·82 − 7·32 or equivalent | M1 | Check diagram for answers. Note: 190·44 – 53·29 = 137·15 | |||||||||
| (\(BD\) = ) \(\sqrt{13.8^2 - 7.3^2}\) or equivalent | m1 | Note: (\(BD\) =) √137·15 FT √’their 137·15’ for m1 only, provided M1 previously gained. | |||||||||
| (\(BD\) =) 11·7(1…..) (cm) | A1 | CAO. Final answer of \(BD\) = 137·15 is M1m0A0. Accept an answer rounded or truncated to at least 1 decimal place. If √137.15 is used correctly for \(BD\) in subsequent work, then award this A1 retrospectively. An unsupported answer of 11·7(1…..) (cm) is awarded M1m1A1. | |||||||||
| \(y = \sin^{-1}\left(\frac{5.5}{11.7}\right)\) or \(\sin^{-1} \dfrac{5{\cdot}5 \times \sin 90}{11{\cdot}7}\) or equivalent | M2 | Check diagram for answers. FT ‘their stated \(BD\)’ (may be on diagram), provided > 5·5. Award M1 for one of the following:
| |||||||||
| \(y\) = 28(·0..) | A1 | Accept an answer rounded or truncated. An unsupported answer of 28(·0..) is awarded M1m1A1M2A1. Allow correct angles given in radians or gradians:
|
16. Alternative method for first 3 marks
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of a ‘two-step’ method. | M2 | A partial trigonometric method is M0. |
| (\(BD\) =) 11·7(1…..) (cm) | A1 |
16. Alternative method for final 3 marks
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of a ‘two-step’ method. | M2 | A partial trigonometric method is M0. |
| (\(y\) =) 28(·0) | A1 | Allow correct angles given in radians or gradians. |