(a) A cottage is on a bearing of 200° and 110° from Dogbe's and Manu's farms respectively. If Dogbe walked 5 km and Manu 3 km from the cottage to their farm...

Answer Details

(a) The cottage \(C\) is at bearing \(200^{\circ}\) from Dogbe's farm \(D\) and \(110^{\circ}\) from Manu's farm \(M\). Reversing bearings, from the cottage:

• Bearing of \(D\) from \(C=200^{\circ}-180^{\circ}=020^{\circ}\), with \(|CD|=5\) km.
• Bearing of \(M\) from \(C=110^{\circ}+180^{\circ}=290^{\circ}\), with \(|CM|=3\) km.

Angle at the cottage \(=290^{\circ}-020^{\circ}=270^{\circ}\Rightarrow\) reflex; the angle between the two directions is \(360^{\circ}-270^{\circ}=90^{\circ}\).

(i) With \(\angle DCM=90^{\circ}\):

\[|DM|=\sqrt{5^{2}+3^{2}}=\sqrt{34}=5.83\approx\mathbf{5.8\text{ km}}\ (2\text{ s.f.}).\]

(ii) Using coordinates \(D=(1.71,4.70),\ M=(-2.82,1.03)\): \(D\!\to\!M=(-4.53,-3.67)\) points south-west.

\[\tan\alpha=\frac{4.53}{3.67}=1.234\Rightarrow\alpha=51^{\circ};\quad\text{bearing}=180^{\circ}+51^{\circ}=\mathbf{231^{\circ}}.\]

(b) Ladder 10 m, wall \(x\) m, foot distance \((x+2)\) m:

\[x^{2}+(x+2)^{2}=10^{2}\Rightarrow 2x^{2}+4x+4=100\Rightarrow x^{2}+2x-48=0\Rightarrow (x+8)(x-6)=0.\]

Since \(x>0,\ \mathbf{x=6\text{ m}}.\)