Mark Scheme

[Maximum Mark: 5]

(a) Angle of Depression
  • The angle of depression from the drone to \(C\) equals the angle of elevation from \(C\) to the drone (alternate interior angles with parallel horizontals). (1 mark)
  • Therefore the angle of depression is 30. (1 mark)

(b) Horizontal Distance \(CB\)
  • Let \(A\) be the point on the ground directly below the drone. Then \(AD=450\ \text{m}\) and triangles \(CAD\) and \(BAD\) are right-angled. (No mark, setup)
  • From \(C\): \(\tan 30^\circ=\frac{AD}{AC}\Rightarrow AC=\frac{450}{\tan30^\circ}=450\sqrt{3}\approx 779.4\ \text{m}\). (1 mark)
  • From \(B\): \(\tan45^\circ=\frac{AD}{AB}\Rightarrow AB=\frac{450}{1}=450\ \text{m}\). (1 mark)
  • Horizontal distance walked: \(CB=AC-AB=450(\sqrt{3}-1)\approx 329.4 m\). (Accept \(329\text{–}330\ \text{m}\)). (1 mark)
Award full credit for equivalent correct methods. Answers: (a) \(30^\circ\); (b) \(CB=450(\sqrt{3}-1)\approx 329.4\ \text{m}\).
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