Question

Two surveyors A and B left P at the same time.A moved at a speed of 6km/hr on a bearing of 043degrees,while B moved at a speed of 4.5 km/hr on a bearing of 115degrees.calculate the distance between A and B after 4 hours and the Bearing of B from A

Answer 1

Answer on Question #61717 – Math – Trigonometry

Question

Two surveyors A and B left P at the same time. A moved at a speed of $6\,\mathrm{km/hr}$ on a bearing of 043 degrees, while B moved at a speed of $4.5\,\mathrm{km/hr}$ on a bearing of 115 degrees. Calculate the distance between A and B after 4 hours and the bearing of B from A.

1) The distance between A and B

$$OA = 6\,\frac{\mathrm{km}}{\mathrm{h}} \cdot 4h = 24\,\mathrm{km},$$ $$OB = 4.5\,\mathrm{km/h} \cdot 4h = 18\,\mathrm{km},$$ $$\angle AOB = 115{}^{\circ} - 43{}^{\circ} = 72{}^{\circ}.$$

By the cosine theorem,

$$AB^2 = OB^2 + OA^2 - 2OB \cdot OA \cdot \cos \angle AOB;$$ $$AB = \sqrt{OB^2 + OA^2 - 2OB \cdot OA \cdot \cos \angle AOB};$$ $$AB = \sqrt{18^2 + 24^2 - 2 \cdot 18 \cdot 24 \cdot 0.309} = \sqrt{324 + 576 - 266.98} = \sqrt{633,02} = 25.2\,\mathrm{km}$$

The distance between A and B after 4 hours is $25.2\,\mathrm{km}$.

2) The bearing B from A

$$\angle KAO = 90{}^{\circ} - 43{}^{\circ} = 47{}^{\circ}$$

By the sine theorem,

$$\begin{array}{l} \frac{AB}{\sin \angle AOB} = \frac{OB}{\sin \angle OAB} \Rightarrow \sin \angle OAB = \frac{OB}{AB} \cdot \sin \angle AOB; \\ \sin \angle OAB = \frac{18}{25.2} \cdot \sin 72{}^{\circ} = \frac{18}{25.2} \cdot 0.951 = 0.679. \\ \angle OAB = 43{}^{\circ}. \end{array}$$

The bearing B from A:

$$270{}^{\circ} - (47{}^{\circ} + 43{}^{\circ}) = 180{}^{\circ}$$

The bearing B from A is $180{}^{\circ}$.

Answer: 1) $25.2\,\mathrm{km}$; 2) $180{}^{\circ}$.

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