Question

A hiker travels 2.5km due North followed by 4.5km on a bearing 132°. Calculate the distance and bearing of the hiker's final position from his initial position.

Answer 1

Answer on Question #51349 – Math – Vector Calculus

A hiker travels $2.5\mathrm{km}$ due North followed by $4.5\mathrm{km}$ on a bearing $132{}^{\circ}$. Calculate the distance and bearing of the hiker's final position from his initial position.

Solution

Fig.1

We need to find the angle $\angle BAC$ and distance between points B and C (see Fig.1).

From the figure 1 it is clear that angle $\angle ABC = 180{}^{\circ} - 132{}^{\circ} = 48{}^{\circ}$.

According to the law of cosines, $(AC) = \sqrt{(AB)^2 + (BC)^2 - 2(AB)(BC)\cos(\angle ABC)}$

$= \sqrt{2.5^2 + 4.5^2 - 2 \cdot 2.5 \cdot 4.5 \cos 38{}^\circ} = 2.96\mathrm{km}$.

According to the law of sines,

$$\frac{BC}{\sin(\angle BAC)} = \frac{AC}{\sin(\angle ABC)} \Rightarrow \sin(\angle BAC) = \frac{BC}{AC} \sin(\angle ABC) \Rightarrow \angle BAC = \arcsin\left(\frac{BC}{AC} \sin(\angle ABC)\right)$$

$\angle BAC = \arcsin\left(\frac{BC}{AC} \sin(\angle ABC)\right) = \arcsin\left(\frac{4.5}{2.96} \sin 38{}^\circ\right) \approx 69{}^\circ$, where $\arcsin(x)$ is the inverse of sine function $\sin(x)$.

Answer: the distance is $2.96\mathrm{km}$ and bearing of the hiker's final position from his initial position is $69{}^{\circ}$.

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