Higher June 2017 Paper 3 Q25
25 Rectangle ABCD is the horizontal base of a triangular prism ABCDEF.
AE = BE
E is vertically above M, the midpoint of AB.
AB = 16 cm AE = 17 cm BC = 30 cm
(a) Show that EM = 15 cm [2 marks]
(b) Work out the size of angle ECM. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(17^2 - (16 \div 2)^2\) or \(\;17^2 - 8^2\) or \(\;289 - 64\) | M1 | Correct use of Pythagoras’ theorem eg \(8^2 + 15^2 = 17^2\) or \(64 + 225 = 289\) |
| \(\sqrt{17^2 - (16 \div 2)^2} \;\; (= 15)\) or \(\sqrt{17^2 - 8^2} \;\; (= 15)\) or \(\sqrt{289 - 64} \;\; (= 15)\) | A1 | Correct use of Pythagoras’ theorem using a square root |
| Alternative method 2 | ||
| \(\sin E = \dfrac{8}{17}\;\) or \(\;\cos A = \dfrac{8}{17}\) or \(E = 28.(\ldots)\) or \(A = 61.9(\ldots)\) or 62 and \(\cos 28.(\ldots) = \dfrac{EM}{17}\) or \(\tan 28.(\ldots) = \dfrac{8}{EM}\) or \(\sin 61.9(\ldots) = \dfrac{EM}{17}\) or \(\tan 61.9(\ldots) = \dfrac{EM}{8}\) | M1 | |
| \(17 \cos 28.(\ldots)\) or \(8 \div \tan 28.(\ldots)\) or \(17 \sin 61.9(\ldots)\) or \(8 \tan 61.9(\ldots)\) | A1 | |
Additional guidance
| 8, 15, 17 on their own | M0A0 |
| \(EM^2 = 289 - 64 = 225,\ EM = 15\) | M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(30^2 + (16 \div 2)^2\) or \(\;30^2 + 8^2\) or 964 | M1 | oe |
| \(\sqrt{\text{their } 964}\) or \(2\sqrt{241}\) or [31, 31.1] | M1dep | oe CM |
| \(\tan x = \dfrac{15}{\text{their } [31,\ 31.1]}\) | M1dep | oe \(\;\) eg \(\;90 - \tan^{-1} \dfrac{\text{their } [31,\ 31.1]}{15}\) dep on M1 M1 |
| [25.7, 26] | A1 | |
| Alternative method 2 | ||
| \(30^2 + 17^2\) \(\;\) or \(\;\) 1189 | M1 | oe |
| \(\sqrt{\text{their } 1189}\) or [34.4, 34.5] | M1dep | oe CE |
| \(\sin x = \dfrac{15}{\text{their } [34.4,\ 34.5]}\) | M1dep | oe \(\;\) eg \(\;90 - \cos^{-1} \dfrac{15}{\text{their } [34.4,\ 34.5]}\) or \(\dfrac{\sin x}{15} = \dfrac{\sin 90}{\text{their } [34.4,\ 34.5]}\) dep on M1 M1 |
| [25.7, 26] | A1 | |
| Alternative method 3 | ||
| \(30^2 + (16 \div 2)^2\) or 964 or \(\;30^2 + 17^2\) or 1189 | M1 | oe |
| \(\sqrt{\text{their } 964}\) or \(2\sqrt{241}\) or [31, 31.1] or \(\sqrt{\text{their } 1189}\) or [34.4, 34.5] | M1dep | oe CM CE |
| \(\cos x = \dfrac{\text{their } [31,\ 31.1]}{\text{their } [34.4,\ 34.5]}\) | M1dep | oe \(\;\) eg \(\;90 - \sin^{-1} \dfrac{\text{their } [31,\ 31.1]}{\text{their } [34.4,\ 34.5]}\) dep on M1 M1 |
| [25.7, 26] | A1 | |
| Alternative method 4 | ||
| \(17^2 - (16 \div 2)^2\) or 225 or \(30^2 + (16 \div 2)^2\) or 964 or \(30^2 + 17^2\) or 1189 | M1 | oe \(EM^2\) \(CM^2\) \(CE^2\) |
| \(\cos x =\) \(\dfrac{\text{their } 964 + \text{their } 1189 - \text{their } 225}{2 \times \sqrt{\text{their } 964} \times \sqrt{\text{their } 1189}}\) | M1dep | oe |
| \(\cos^{-1} \dfrac{\text{their } 964 + \text{their } 1189 - \text{their } 225}{2 \times \sqrt{\text{their } 964} \times \sqrt{\text{their } 1189}}\) | M1dep | oe dep on M1 M1 |
| [25.7, 26] | A1 | |