4MA0 Higher June 2016 Paper 4R Q11
11 Here is a triangle \(QRS\).
Diagram NOT accurately drawn
\(SQ\) = 15 cm
Angle \(RSQ\) = 90°
Area of triangle \(QRS\) = 60 cm²
Work out the size of angle \(SQR\).
Give your answer correct to 1 decimal place.
(4)
| Scheme | Marks |
|---|---|
| \(SR = (60 \div 15) \times 2\) (=8) | M1 |
| \(\tan SQR = \dfrac{\text{‘8’}}{15}\) | M1ft |
| \(SQR = \tan^{-1}\left(\dfrac{\text{‘8’}}{15}\right)\) | M1ft |
| 28.1 | A1 |
| (4) |
Notes
M1ft: (or M1 for \(\sin SQR = \dfrac{\text{‘8’}}{\text{‘17’}}\) or \(\cos SQR = \dfrac{15}{\text{‘17’}}\) where ‘17’ comes from a fully correct method)
M1ft: (or \(\sin^{-1}\left(\dfrac{\text{‘8’}}{\text{‘17’}}\right)\) or \(\cos^{-1}\left(\dfrac{15}{\text{‘17’}}\right)\))
A1: 28.07 – 28.1