Higher June 2023 Paper 6 Q8
8 Taylor designs a logo using isosceles triangles joined at a central point, P.
This is the start of Taylor’s design.
Not to scale
The completed design will have rotational symmetry, order 60 about point P.
Each triangle has base, \(b\), and height, \(h\), measured in mm.
Not to scale
Calculate \(h\) when \(b = 40\) mm.
Give your answer correct to 1 decimal place. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.1[0…] nfww | 4 | M1 for \(\frac{360}{60}\) oe soi by 6 AND Method 1 using tan: M2 for [\(h\) = ] 20 tan(their 6) oe or M1 for correct use of tan(their 6) oe or Method 2 using sine rule: M2 for [\(h\) = ] \(\frac{20 \sin(their\ 6)}{\sin(90 - their\ 6)}\) or M1 for \(\frac{\sin(their\ 6)}{h} = \frac{\sin(90 - their\ 6)}{20}\) oe or Method 3 using cos and Pythagoras: M2 for \(\sqrt{\left(\frac{20}{\cos(their\ 6)}\right)^2 - 20^2}\) or M1 for \(\left(\frac{20}{\cos(their\ 6)}\right)^2 - 20^2\) | May be on diagram In all methods, if their angle is not 6 then method must be seen, not implied by interim answers unless stated otherwise Accept any acute angle used for their 6 eg [\(h\) = ] \(\frac{20}{\tan(90 - their\ 6)}\) eg tan(their 6) = \(\frac{h}{20}\) NBs \(\frac{\text{approx. circumference}}{60} = \frac{40\pi}{60}\) = 2.1 scores 0 20sin6 = 2.1 scores M1 for 6 Solution from scale drawing scores a maximum of M1 if 6 seen |