Bearings Question involving cosine rule

trigonometry

Ship A is 120 nautical miles from lighthouse L on a bearing of 072°T, while ship B is 180 nautical miles from L on a bearing of 136°T. Calculate the distance between the two ships to the nearest nautical mile.

I've stuck on this question for a while.I have tried using the cosine rule, where I construct a triangle in which the angle is 136°, side a is 120 nautical miles side b is 180 nautical miles. But I'm nowhere near the answer even after doing that. I get the answer 279 for the missing side after calculating

$\sqrt{180^2+120^2-2*180*120*\cos(136)}$

What am I doing wrong?

Best Answer

In general: to find the length of a side of the triangle by a cosine rule the following equation applies: $a^2 = b^2 + c^2 - 2*b*c*cos(A)$, where "a", "b" and "c" are the sides of your triangle and "A" is the angle opposite to your "a" side. So in order to evaluate "a" you need to square root the right side of your equation. Can you attach a figure showing your problem?

Related Solutions

[Math] Trigonometry Airplane question. Finding bearing and distance.

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
|   /
|  /
| /
|/
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: $$\theta=\tan^{-1}\left(\frac{AC}{AB+480}\right) = \tan^{-1}\left(\frac{640\sin(59^\circ)}{640\cos(59^\circ)+480}\right) \approx 34^\circ \text{ to get the bearing you subtract that value from 90 to get } S56^\circ W$$

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Related Solutions

[Math] Trigonometry Airplane question. Finding bearing and distance.

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