Numeracy Higher June 2022 Unit 2 Q10
10. Rhian is playing golf.
The diagram below shows one of the holes she is playing.
The dotted lines show the recommended route to the hole, taking 2 shots.
The length of a shot is the straight-line distance as measured on horizontal ground.
The recommended first shot is 160 yards. This shot is to a point equidistant from the sand and the tree on the corner. It is then followed by a second shot of 148 yards to the hole.
Diagram not drawn to scale
Diagram not drawn to scale
It shows an arc of a circle of length 65 yards, with its centre at the start.
Rhian does not want her first shot to land in the trees or in the sand.
The greatest angle she can hit her first shot to the right or left of the recommended line has been shown on the diagram as \(x\).
Calculate the size of angle \(x\), giving your answer correct to 3 significant figures. [4]
To get over the trees, this shot needs to be longer than 160 yards.
Diagram not drawn to scale
Calculate the length of this single shot. [7]
From the formula list: In any triangle \(ABC\): Sine rule \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\); Cosine rule \(a^2 = b^2 + c^2 - 2bc\cos A\); The solutions of \(ax^2 + bx + c = 0\) where \(a \neq 0\) are given by \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
| Scheme | Marks | Comments |
|---|---|---|
| Sight of (2 ×) \(\dfrac{x}{360} \times 2 \times \boldsymbol{\pi} \times 160\) or equivalent | B1 | |
| (x =) \(\dfrac{65 \div 2 \times 360}{2 \times \boldsymbol{\pi} \times 160}\) or equivalent | M1 | Allow for sight of \(\dfrac{65 \times 360}{2 \times \boldsymbol{\pi} \times 160}\) or equivalent |
| = 11.6 (º) | A2 | CAO A1 for 11.6366… to 11.6441… or 585/16π OR A1 for a final answer of 23.3(º) from failing to halve their sector angle If no marks awarded, SC1 for a final answer of 23.3(º) from use of diameter 160 yards and halving their sector angle |
Corrected by Westie's Workshop
- The board's mark scheme printed “use of diameter 160 cm”. Changed to “use of diameter 160 yards” because all the distances in this question are in yards, not centimetres.
Original: see the board's mark scheme PDF (link above).
| Scheme | Marks | Comments |
|---|---|---|
| (Angle at hole =) \(\sin^{-1}\!\left(\dfrac{\sin 32}{148} \times 160\right)\) | M2 | M1 for \(\dfrac{\sin(\text{Angle at hole})}{160} = \dfrac{\sin 32}{148}\) or equivalent |
| (Angle at hole =) 34.9(5…) or 35 (º) | A1 | CAO |
| (3rd Angle = 180–32–34.9(5…) =) 113(.048…) (º) | B1 | FT 180 – 32 – ‘their 34.9(5…)’ provided sine rule attempted |
| (Length of single shot =) \(\dfrac{148}{\sin 32} \times \sin 113(.048\ldots)\) OR \(\dfrac{160}{\sin 34.9(5\ldots)} \times \sin 113(.048\ldots)\) OR \(\sqrt{148^2 + 160^2 - 2 \times 148 \times 160 \times \cos\, 113(.048\ldots)}\) | M2 | FT ‘their derived 113.04(8…)’ provided it is clearly their 3rd angle for any of the 3 possible methods M1 for \(\dfrac{\text{single shot}}{\sin 113(.048\ldots)} = \dfrac{148}{\sin 32}\) or equivalent OR FT ‘their derived 34.9(5…)’ provided it is clearly their angle at the hole M1 for \(\dfrac{\text{single shot}}{\sin 113(.048\ldots)} = \dfrac{160}{\sin 34.9(5\ldots)}\) or equivalent OR M1 for shot2 = 1482+1602−2×148×160×cos113(.048…) |
| (= \(\sqrt{66045(.770\ldots}\) ) = 256.7 to 257.1 (yards) | A1 |
Alternative
10(b) Alternative method 1:
| Scheme | Marks | Comments |
|---|---|---|
 | Only award these marks if a clear intention made to split the triangle this way, and a full method attempted to find the length of the single shot Allow answers to be suitably rounded Only penalise the final A1 mark if their answer from using rounded values does not lie in the range given | |
| (x =) 160 × sin 32 = 84.7(787…) or 84.8 (yards) | M1 A1 | |
| (y =) 160 × cos 32 OR \(\sqrt{160^2 - 84.7(787\ldots)^2}\) | M1 | FT ‘their 84.7(787…)’ provided 1st M1 awarded |
| = 135.6(876…) or 135.7 (yards) | A1 | |
| (z =) \(\sqrt{148^2 - 84.7(787\ldots)^2}\) | M1 | Or a full alternative method FT ‘their 84.7(787…)’ provided 1st M1 awarded |
| = 121.3(060…) (yards) | A1 | |
| (Length of shot = 135.6(876..) + 121.3(060…)=) 256.8 to 257.15 (yards) | A1 | FT ‘their 135.6(876…)’ and ‘their 121.3(060…)’ provided M1M1M1 previously awarded and at least one A1 previously awarded |
Alternative
10(b) Alternative method 2:
| Scheme | Marks | Comments |
|---|---|---|
| Strategy to form a quadratic equation and solve | S1 | |
| 1482 = 1602 + shot2 – 2×160×shot×cos32 | M1 | |
| Shot2 – 320cos32 × shot + 3696 = 0 | A1 | Note: 320cos32 = 271.375… |
| (Shot =) \(\dfrac{271.375\ldots \pm \sqrt{271.375^2 - 4 \times 1 \times 3696}}{2 \times 1}\) | M1 | FT ‘their 320cos32’ Must be seen |
| (Shot =) \(\dfrac{271.375\ldots \pm \sqrt{58860.60272}}{2}\) | A1 | |
| (Shot =) 256.99… or 257 (yards) ( or 14.38… (yards)) | A2 | Implies previous A1 A1 if 256.99… clearly not identified as being their answer |
Corrected by Westie's Workshop
- In Alternative methods 1 and 2 the board's mark scheme printed the unit “(m)” for the lengths 84.8, 135.7, 121.3, 257 and 14.38…. Changed to “(yards)” because all the distances in this question are in yards, not metres (the main method and the final answer line already say yards).
Original: see the board's mark scheme PDF (link above).