trigonometry
I am having problem solving the following question related to "bearing" in trigonometry.
A motorboat is 10 km South of a lighthouse and is on a course of 053 degrees. What is the shortest distance between the motorboat and the lighthouse?
My confusion is how does a boat travel 10 km South of a lighthouse and making an angle of 53 degrees clockwise from North. Is there something wrong with the question itself?
Thanks in advance.
Cosine Law would be easier, but let's do it your way. Consider the small right triangle (call it $\Delta ABC$), which roughly looks like:
A----B
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C
Note that $\angle ABC =59^\circ$. Thus, $AB=640\cos(59^\circ)$ and $AC=640\sin(59^\circ)$.
Now consider the large right triangle. Its two legs are $AC$ and $AB+480$. Hence, by the Pythagorean Theorem, we have: $$x=\sqrt{(AC)^2 + (AB+480)^2}=\sqrt{(640\cos(59^\circ)+480)^2 + (640\sin(59^\circ))^2} \approx 978 \text{ miles}$$ Finally, for the angle bearing: $$\theta=\tan^{-1}\left(\frac{AC}{AB+480}\right) = \tan^{-1}\left(\frac{640\sin(59^\circ)}{640\cos(59^\circ)+480}\right) \approx 34^\circ \text{ to get the bearing you subtract that value from 90 to get } S56^\circ W$$
I assume that the angles are measured in a clockwise direction starting from a reference direction of $0^\circ$ which is north.
Using trigonometry the problem can be reduced to find an angle in a triangle knowing two of its sides and another angle.
The water velocity is represented by the blue vector, with a magnitude of $3\, \mathrm{km/h}$. To answer the problem we need to find the direction of the black vector, whose magnitude is $35\, \mathrm{km/h}$. This means we need to find the angle $\theta=A$ of the triangle $\Delta=[A,B,C]$, knowing the sides $a=3$ and $b=35$, and the angle $B=105^\circ$. From the law of sines
$$\frac{a}{\sin A}=\frac{b}{\sin B}$$
we deduce that
$$\sin\theta=\sin A=\frac{a\sin B}{b}=\frac{3\sin 105^\circ}{35}.$$
So
$$\theta=\arcsin\left( \frac{3\sin 105^\circ}{35}\right)\approx 4.749^\circ .$$
This means that the boat should take the direction of
$$65^\circ+\theta\approx 69.75^\circ,$$
sailing with a speed of $35\, \mathrm{km/h}$. A new application of the law of sines to the side $c$ and angle $C$, yields $c=34.103\, \mathrm{km/h}$, which is the resulting speed of the boat in the port direction.
Best Answer
Here is a drawing of the problem I made (Excuse my bad drawing skills).
LH represents the lighthouse, and MB represents the motorboat.
Here, the motorboat is travelling at a $53^{\circ}$ angle from North.
Now, here is the important part to notice: The shortest distance will be the point when the perpendicular of the black line (The path which the motorboat goes) intersects the light house.
Therefore, the $\color{red}{\text{red}}$ line is the shortest distance. This length can easily be calculated using trigonometry:
$10\sin{53^{\circ}}\approx \boxed{7.986 \text{ km}}$.