Higher January 2022 Paper 1 Q21
21 The diagram shows the prism \(ABCDEFGHJK\) with horizontal base \(AEFG\)
Diagram NOT accurately drawn
\(ABCDE\) is a cross section of the prism where
\(ABDE\) is a square
\(BCD\) is an equilateral triangle
\(EF = 2 \times AE\)
\(M\) is the midpoint of \(GF\) so that \(JM\) is vertical.
Angle \(MAJ = y^\circ\)
Given that \(\tan y^\circ = T\)
find the value of \(T\), giving your answer in the form \(\dfrac{\sqrt{p} + \sqrt{q}}{17}\) where \(p\) and \(q\) are integers.
(5)
| Scheme | Marks |
|---|---|
eg \((AM =)\;\sqrt{x^2 + (4x)^2}\;\left(= \sqrt{17x^2} = x\sqrt{17}\right)\) oe or \((AM =)\;\sqrt{(0.5x)^2 + (2x)^2}\;\left(= \sqrt{\dfrac{17}{4}x^2} = \dfrac{x\sqrt{17}}{2}\right)\) oe or \((AM =)\;\sqrt{20^2 + 5^2}\;\left(= \sqrt{425} = 5\sqrt{17}\right)\) oe | M1 |
Height of triangle eg \(\sqrt{(2x)^2 - x^2}\;\left(= \sqrt{3x^2} = x\sqrt{3}\right)\) oe or \(\sqrt{x^2 - (0.5x)^2}\;\left(= \sqrt{\dfrac{3}{4}x^2} = \dfrac{x\sqrt{3}}{2}\right)\) oe or \(\sqrt{10^2 - 5^2}\;\left(= \sqrt{75} = 5\sqrt{3}\right)\) oe | M1 |
| eg \(\tan MAJ = \dfrac{\sqrt{3} + 2}{\sqrt{17}}\) or \(\tan MAJ = \dfrac{\frac{\sqrt{3}}{2} + 1}{\frac{\sqrt{17}}{2}}\) or \(\tan MAJ = \dfrac{5\sqrt{3} + 10}{5\sqrt{17}}\) | M1 |
| eg \(\dfrac{(\sqrt{3} + 2)}{\sqrt{17}} \times \dfrac{\sqrt{17}}{\sqrt{17}}\left(= \dfrac{\sqrt{51} + 2\sqrt{17}}{17}\right)\) | M1 |
| \(\dfrac{\sqrt{68} + \sqrt{51}}{17}\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for a correct method to find \(AM\) as a numerical value or in algebraic form, must have brackets or recover
M1: for a correct method to find height of equilateral triangle \(HJK\) as a numerical value or in algebraic form
M1: for correct values for the correct angle (no algebra) or for \(\tan MAJ\) is given numerically in the range 0.9 – 0.91
A1: or \(\dfrac{\sqrt{51} + \sqrt{68}}{17}\)