Higher June 2019 Paper 2 Q19
19 The diagram shows a triangular prism.
The base, \(ABCD\), of the prism is a square of side length 15 cm.
Angle \(ABE\) and angle \(CBE\) are right angles.
Angle \(EAB = 35^\circ\)
\(M\) is the point on \(DA\) such that
\(DM : MA = 2 : 3\)
Calculate the size of the angle between \(EM\) and the base of the prism.
Give your answer correct to 1 decimal place. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 31.0 | P1 | for \(\tan 35 = BE \div 15\) or \(BE = 10.5(0\ldots)\) OR finding the length \(DM = \dfrac{2}{5} \times 15\ (= 6)\) or \(MA = \dfrac{3}{5} \times 15\ (= 9)\) or 6 : 9 OR showing the required angle on a diagram, eg with an arc |
| P1 | for \(MB = \sqrt{15^2 + \text{``}9\text{''}^2}\) or \(\sqrt{306}\) or 17.4(9…) OR \(ME = \sqrt{\text{``}9\text{''}^2 + \text{``}18.3(1\ldots)\text{''}^2}\) or \(\sqrt{416.(3\ldots)}\) or 20.4(0…) | |
| P1 | for using an appropriate trigonometric ratio to set up an equation in angle \(EMB\) eg \(\tan\theta = \text{``}10.5(0\ldots)\text{''} \div \text{``}17.4(9\ldots)\text{''}\) or \(\cos\theta = \text{``}17.4(9\ldots)\text{''} \div \text{``}20.4(0\ldots)\text{''}\) or \(\sin\theta = \text{``}10.5(0\ldots)\text{''} \div \text{``}20.4(0\ldots)\text{''}\) | |
| A1 | for answer in the range 30.9 to 31 |
Additional guidance
\(MB = \sqrt{9^2 + 15^2} = \sqrt{306}\) (= 17.4(9…) or 17.5)
\(BE = 15 \times \tan 35\) (= 10.5(0…))
\(AE = 15 \div \cos 35\) (= 18.3(1…))
\(ME = \sqrt{9^2 + 18.31\ldots^2} = \sqrt{416.(3\ldots)}\) (= 20.4(0…))
Check diagram for working
If an answer is shown in the range in working and then incorrectly rounded award full marks.