Higher June 2023 Paper 3 Q23
23
Not drawn accurately
A boat sails 35 km North from \(A\) to \(B\).
From \(B\) the boat sails to \(C\) and then back to \(A\).
(a) Show that the distance the boat sails from \(C\) to \(A\) is 79 km to the nearest km
You must show your working. [2 marks]
(b) Work out the bearing of \(A\) from \(C\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(35^2 + 65^2 - 2 \times 35 \times 65 \times \cos 100\) | M1 | oe valid trigonometric method used must be correct |
| \(\sqrt{35^2 + 65^2 - 2 \times 35 \times 65 \times \cos 100} = 78.9(\ldots)\) or \(\sqrt{6240.(0992\ldots)} = 78.9(\ldots)\) | A1 | \(CA = 78.99429858\) |
Additional guidance
| Using sine rule with \(CA = 79\) to obtain \(AB\) or \(BC\) | M0A0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – sine rule to find \(ACB\) | ||
| \(\dfrac{\sin ACB}{35} = \dfrac{\sin 100}{79}\) | M1 | oe 79 may be 78.9(…) |
| \(\sin ACB = 35 \times \dfrac{\sin 100}{79}\) or \(\sin ACB = 35 \times 0.0124\ldots\) or \(\sin ACB = 0.436\ldots\) | M1dep | oe |
| \(ACB =\) [25.8, 26] | A1 | |
| 234.(…) | A1ft | ft \(360 - 100 -\) their \(ACB\) with M2 scored |
| Alternative method 2 – cosine rule to find \(ACB\) | ||
| \(35^2 = 79^2 + 65^2 - 2 \times 79 \times 65 \times \cos ACB\) | M1 | oe 79 may be 78.9(…) |
| \(\cos ACB = \dfrac{79^2 + 65^2 - 35^2}{2 \times 79 \times 65}\) or \(\cos ACB = \dfrac{9241}{10\,270}\) or \(\cos ACB = 0.899\ldots\) | M1dep | |
| \(ACB =\) [25.8, 26] | A1 | |
| 234.(…) | A1ft | ft \(360 - 100 -\) their \(ACB\) with M2 scored |
| Alternative method 3 – sine rule to find \(BAC\) | ||
| \(\dfrac{\sin BAC}{65} = \dfrac{\sin 100}{79}\) | M1 | oe 79 may be 78.9(…) |
| \(\sin BAC = 65 \times \dfrac{\sin 100}{79}\) or \(\sin BAC = 65 \times 0.0124\ldots\) or \(\sin BAC = 0.81(0\ldots)\) | M1dep | oe |
| \(BAC =\) [54.1, 54.3] | A1 | |
| 234.(…) | A1ft | ft their \(BAC + 180\) with M2 scored |
| Alternative method 4 – cosine rule to find \(BAC\) | ||
| \(65^2 = 79^2 + 35^2 - 2 \times 79 \times 35 \times \cos BAC\) | M1 | oe 79 may be 78.9(…) |
| \(\cos BAC = \dfrac{79^2 + 35^2 - 65^2}{2 \times 79 \times 35}\) or \(\cos BAC = \dfrac{3241}{5530}\) or \(\cos BAC = 0.586\ldots\) | M1dep | |
| \(BAC =\) [54.1, 54.3] | A1 | |
| 234.(…) | A1ft | ft their \(BAC + 180\) with M2 scored |
Additional guidance
\(CA = 79\) is given in part (a) or 78.9(…) can be used. There is no follow through from part (a).
Accept any notation for the angle eg \(\sin x\) or \(\sin C\) for angle \(ACB\)
Correct work for part (b) seen in part (a) may be awarded method marks in part (b)