Question

A boat travels north at a speed of 20 mph and a bearing of N 32 degrees E. Another boat travels at a speed of 28 mph and a bearing of S 42 degrees E. After 2 hours, how far apart are the boats

Answer 1

• 78.23 miles

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It is assumed boats leave from the same point.

Boats travel after 2 hours:

• 20*2 = 40 miles and 28*2 = 56 miles.

The angle formed between the two directions:

• 90° - 32° + 90° - 42° = 108°

The line between the endpoints is opposite to this angle.

Use the law of cosines and find the distance:

• 
  `d=√(40^2+56^2-2*40*56*cos 108) \approx 78.23 \ miles`

Answer 2

77.3 miles apart (nearest tenth)

Explanation:

Given information:

• Boat A travels north at a speed of 20 mph and a bearing of N32°E.
• Boat B travels at a speed of 28 mph and a bearing of S42°E.
• After 2 hours, Boat A will have travelled 40 miles.
• After 2 hours, Boat B will have travelled 56 miles.

Draw a diagram using the given information (see attached).

To find how far apart the boats are, model as a triangle and find the length of the missing side by using the cosine rule.

`\boxed{\begin{minipage}{6 cm}\underline{Cosine Rule} \\\\$c^2=a^2+b^2-2ab \cos C$\\\\where:\\ \phantom{ww}$\bullet$ $a, b$ and $c$ are the sides.\\ \phantom{ww}$\bullet$ $C$ is the angle opposite side $c$. \\\end{minipage}}`

From inspection of the drawn diagram:

• a = 40 miles
• b = 56 miles
• c = distance between the boats
• C = 180° - 32° - 42° = 106°

Substitute the values into the cosine rule and solve for c:

`\implies c^2=40^2+56^2-2(40)(56) \cos 106^(\circ)`

`\implies c^2=1600+3136-4480 \cos 106^(\circ)`

`\implies c^2=4736-4480 \cos 106^(\circ)`

`\implies c=\sqrt{4736-4480 \cos 106^(\circ)}`

`\implies c=77.27131004`

Therefore, the boats are 77.3 miles apart (nearest tenth) after 2 hours.