Question

Two ranger stations located 10 km apart receive a distress call from a camper. Electronic equipment allows them to determine that the camper is at an angle of 71 degrees from the first station and 100 degrees from the second, each angle having as one side the line of the stations. Find the distance from the camper to each station?

Answer 1

Solution

We first draw the diagram of the problem

From the diagram above,

A is used to denote the first station;

B is used to denote the second station

C is used to denote the camper location

x is the distance between the first station and the camper

y is the distance between the second station and the camper

Therefore, we want to find x and y

We complete the above triangle by finding the remaining two angles

Using the Sine Rule

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

From the triangle above

$\begin{gathered} \frac{x}{\sin80}=\frac{100}{\sin 29} \\ \text{cross multiply} \\ x\times\sin 29=100\times\sin 80 \\ x\sin 29=100\sin 80 \\ x=\frac{100\sin 80}{\sin 29} \\ x=203.1328818 \\ x=203.13\operatorname{km} \end{gathered}$

Thus, the distance of the camper from the first station is 203.13km

Now, we are left with finding the distance of the camper from the second station

$\begin{gathered} \frac{y}{\sin71}=\frac{100}{\sin 29} \\ \text{cross multiply} \\ y\times\sin 29=100\times\sin 71 \\ y\sin 29=100\sin 71 \\ y=\frac{100\sin 71}{\sin 29} \\ y=195.0288394 \\ y=195.03\operatorname{km} \end{gathered}$

Therefore, the distance of the camper from the second station is 195.03km