Intermediate June 2025 Unit 2 Q15
15.
(a) Calculate the length of PQ. [3]
Diagram not drawn to scale
(b) Calculate the size of angle \(x\). [3]
Diagram not drawn to scale
| Scheme | Marks | Comments |
|---|---|---|
| (PQ2 = ) 19·32 – 16·82 or equivalent | M1 | Check diagram for answers. (PQ2 =) 372·49 – 282·24 (ignore place values for M1) = 90·25 |
| (PQ =) \(\sqrt{19 \cdot 3^2 - 16 \cdot 8^2}\) or equivalent | m1 | (PQ =) √90·25 Sight of √‘their 90·25’ or √‘their 90·25’ evaluated is awarded m1 provided M1 previously gained. |
| (PQ =) 9·5 (cm) | A1 | CAO. Mark final answer. Allow 9 or 10 provided 9·5 has been seen. Final answer of PQ = 90·25 is M1m0A0. An unsupported correct answer is awarded M1m1A1. |
15.(a) Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of a two-step trigonometric method | M2 | A partial trigonometric method is awarded M0. |
| (PQ =) 9·5 (cm) | A1 | CAO. Mark final answer. Allow 9 or 10 provided 9·5 has been seen. |
| Scheme | Marks | Comments | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| (\(x\) =) \(\sin^{-1} \dfrac{15{\cdot}4}{20{\cdot}5}\) or \(\sin^{-1} \dfrac{15{\cdot}4 \times \sin 90}{20{\cdot}5}\) or equivalent | M2 | Check diagram for answers. M1 for any one of the following:
| |||||||||
| Correct evaluation in the range 48·6 to 48·7(°) | A1 | Mark final answer. Allow 49 from correct working. Allow correct angles given in radians or gradians.
|
15.(b) Alternative method:
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of ‘two-step’ method. | M2 | A partial trigonometric method is M0. |
| Correct evaluation in the range 48·6 to 48·7(°). | A1 | Mark final answer. Allow 49 from correct working. Allow correct angles given in radians or gradians. |