Higher June 2019 Paper 3 Q23
23 The diagram shows the positions of three towns, Acton (\(A\)), Barston (\(B\)) and Chorlton (\(C\)).
Barston is 8 km from Acton on a bearing of 037°
Chorlton is 9 km from Barston on a bearing of 150°
Find the bearing of Chorlton from Acton.
Give your answer correct to 1 decimal place.
You must show all your working. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 098.6 | P1 | for using bearings to determine \(ABC\) as 67° |
| P1 | for using the cosine rule to find \(AC\) eg (\(AC^2 =\)) \(9^2 + 8^2 - 2 \times 9 \times 8 \times \cos[67]\) oe or \(AC = 9.4199\ldots\) | |
| P1 | (dep P1) for using the sine rule to find angle \(BAC\) eg \(\dfrac{9}{\sin BAC} = \dfrac{\text{``}9.42\text{''}}{\sin[67]}\) oe OR for using the cosine rule to find angle \(BAC\) eg \(9^2 = \text{``}9.42^2\text{''} + 8^2 - 2 \times \text{``}9.42\text{''} \times 8 \times \cos BAC\) oe | |
| P1 | for rearranging eg \(\sin BAC = 9 \times \dfrac{\sin[67]}{\text{``}9.42\text{''}}\) oe OR eg \(\cos BAC = (\text{``}9.42^2\text{''} + 8^2 - 9^2) \div (2 \times \text{``}9.42\text{''} \times 8)\) oe OR for angle \(BAC = 61.57\ldots\) | |
| A1 | for angle in the range 98.5 to 98.6 |
Additional guidance
Accept 67 written on the diagram.
Accept correct substitution into RHS of equation
Accept \(AC\) in the range 9.41 to 9.42
Accept any equivalent form with values substituted
If the correct answer is given without supportive evidence award 0 marks.
Condone missing “0” at the front.
If an answer within the range is seen in working and rounded incorrectly award full marks.