Higher June 2018 Paper 2 Q19
19 The diagram shows a triangular prism.
Diagram NOT accurately drawn
\(AF\) = 10 cm, \(AB\) = 24 cm and \(BC\) = 8 cm.
Angle \(FAB\) = angle \(ADC\) = angle \(BCD\) = 90°
Work out the size of the angle between the line \(BE\) and the plane \(ABCD\).
Give your answer correct to 1 decimal place.
(3)
| Scheme | Marks |
|---|---|
\(BE^2 = 10^2 + 24^2 + 8^2\) (= 100 + 576 + 64 = 740) (\(BE = 2\sqrt{185}\) = 27.202 …) or \(BD^2 = 8^2 + 24^2\) (= 64 + 576 = 640) (\(BD = 8\sqrt{10}\) = 25.298….) | M1 |
\(\sin DBE = \dfrac{10}{\sqrt{\text{``}{740}\text{''}}}\) (= 0.3676 …) or \(\tan DBE = \dfrac{10}{\sqrt{\text{``}{640}\text{''}}}\) (= 0.3952…) or \(\cos DBE = \dfrac{\sqrt{\text{``}{640}\text{''}}}{\sqrt{\text{``}{740}\text{''}}}\) (=0.9428…) | M1 |
| 21.6 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: Complete method to find \(BE\) or \(BE^2\) or \(BD\) or \(BD^2\)
M1: Allow use of sine or cosine rule
\(\sin DBE = \dfrac{10\sin 90}{\sqrt{\text{``}{740}\text{''}}}\) or
\(\cos DBE = \dfrac{\text{``}{640}\text{''} + \text{``}{740}\text{''} - 10^2}{2 \times \sqrt{\text{``}{640}\text{''}} \times \sqrt{\text{``}{740}\text{''}}}\)
(=0.9299...)
A1: 21.5 – 21.6