Question

In an orienteering race the first checkpoint is 500m from the start and on a bearing of 030°. The second checkpoint is 400m from the first checkpoint and 600m from the start. Find, to the nearest degree, the two possible bearings of checkpoint 2 from the start.

Answer 1

Answer on Question #51337 – Math – Vector Calculus

In an orientering race the first checkpoint is $a = 500\mathrm{m}$ from the start and on a bearing of $\varphi = 30{}^{\circ}$. The second checkpoint is $b = 400\mathrm{m}$ from the first checkpoint and $l = 600\mathrm{m}$ from the start. Find, to the nearest degree, the two possible bearings $\alpha_{1}$ and $\alpha_{2}$ of checkpoint 2 from the start.

Solution

$\alpha_{1}$:

According to the law of cosines used in triangle 102, where 1 is the first checkpoint, 0 is the start, 2 is one of possible second checkpoints, obtain

$$b^{2} = a^{2} + l^{2} - 2al \cos(\alpha_{1} - \varphi)$$ $$\begin{array}{l} \alpha_{1} = \arccos \left(\frac{a^{2} + l^{2} - b^{2}}{2al}\right) + \varphi = \arccos \left(\frac{250000\mathrm{m}^{2} + 360000\mathrm{m}^{2} - 160000\mathrm{m}^{2}}{600000\mathrm{m}^{2}}\right) + 30{}^{\circ} = \\ = \arccos 0.75 + 30{}^{\circ} = 41{}^{\circ} + 30{}^{\circ} = 71{}^{\circ} \end{array}$$

where $\arccos(x)$ is the inverse of cosine function $\cos(x)$.

$\alpha_{2}$:

According to the law of cosines used in triangle $102'$, where 1 is the first checkpoint, 0 is the start, $2'$ is one of possible second checkpoints, obtain

$$b^{2} = a^{2} + l^{2} - 2al \cos(\varphi - \alpha_{2})$$ $$\alpha_{2} = \varphi - \arccos \left(\frac{a^{2} + l^{2} - b^{2}}{2al}\right) = 30{}^{\circ} - \arccos \left(\frac{250000\mathrm{m}^{2} + 360000\mathrm{m}^{2} - 160000\mathrm{m}^{2}}{600000\mathrm{m}^{2}}\right) =$$ $$= 30{}^{\circ} - \arccos 0.75 = 30{}^{\circ} - 41{}^{\circ} = -11{}^{\circ} \text{ or } 169{}^{\circ}$$

**Answer**: $\alpha_{1,2} = \varphi \pm \arccos \left(\frac{a^2 + l^2 - b^2}{2al}\right) = 71{}^{\circ}, -11{}^{\circ}$.

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