The Ambiguous Case of the Sine Law FOM11 4.3 #7

CanadianMathMom

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Can anyone help figure this question out? We have spent over an hour, and can not come up with the correct answer. If you can show (or explain your work), it would be truly appreciated. 11519
 

Jomo

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Welcome to the forum.
Can you please show us the work you did so we can see where you are making a mistake? Please show us a diagram. Thank you.
 

MarkFL

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Hello, and welcome to FMH! :)

Consider the following diagram:

fmh_0026.png

The boats starts at the dock, then moves 5 km up to the buoy, and from there it travels an additional 3 km, which could put it anywhere on the circle, however the angle between the boat and the buoy is \(12^{\circ}\) as seen from the dock, which means the boat must also lie along the dotted line. We see there are two possible locations for the boat.

Can you proceed?
 

CanadianMathMom

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Hello, and welcome to FMH! :)

Consider the following diagram:

View attachment 11520

The boats starts at the dock, then moves 5 km up to the buoy, and from there it travels an additional 3 km, which could put it anywhere on the circle, however the angle between the boat and the buoy is \(12^{\circ}\) as seen from the dock, which means the boat must also lie along the dotted line. We see there are two possible locations for the boat.

Can you proceed?
Good morning! Thank you for taking the time to help us. The answer is 7.7km (distance from current canoe and buoy). I am including our work, and the textbook answer. Im not sure if we are missing a step, or have labelled the diagram incorrect. Thanks again!1152511526
 

Dr.Peterson

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I see two mistakes. First, the 12 degree angle is measured at the dock, not at the turning point. Second, the answer of 7.7 is for option 1, not option 2 (as they assume option 1 is the more natural interpretation). The actual work will be very different from what you show.

By the way, did you notice that, although you stated that the problem illustrates the ambiguous case of the Law of Sines, you used the Law of Cosines, which has no ambiguity? This problem can be solved either by using the Law of Sines, as implied, or by using the Law of Cosines in reverse, which leads to a quadratic equation with two solutions.
 
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CanadianMathMom

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I see two mistakes. First, the 12 degree angle is measured at the dock, not at the turning point. Second, the answer of 7.7 is for option 1, not option 2 (as they assume option 1 is the more natural interpretation). The actual work will be very different from what you show.

By the way, did you notice that, although you stated that the problem illustrates the ambiguous case of the Law of Sines, you used the Law of Cosines, which has no ambiguity? This problem can be solved either by using the Law of Sines, as implied, or by using the Law of Cosines in reverse, which leads to a quadratic equation with two solutions.
Thank you kindly for your help. It is sincerely appreciated. We worked through the problem, and solved with the correct answer. Thanks to all who helped us get through this!! 11529
 

JeffM

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Obviously, answers # 3 and # 5 are correct, but they do ignore one minor issue. The twelve degree angle could lie to the right (as drawn in answer # 3) or to the left. There are FOUR positions where the canoe may be. This does not invalidate the suggested answers of course because the question asks about distances rather than positions, and the distance to the farther position to the right equals (by congruence) the distance to the farther position on the right, and the two nearer distances are equal as well for the same reason.
 

MarkFL

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I would convert angles to radians, and write (using the Law of Sines):

\(\displaystyle \frac{3}{\sin\left(\dfrac{\pi}{15}\right)}=\frac{5}{\sin(\alpha)}\)

And this implies:

\(\displaystyle \alpha=\frac{\pi}{2}\pm\arccos\left(\frac{5}{3}\sin\left(\frac{\pi}{15}\right)\right)\)

Now, we know, since the sum of the 3 interior angles is \(\pi\), that we must have:

\(\displaystyle \frac{\pi}{15}+\alpha+\beta=\pi\)

Hence:

\(\displaystyle \beta=\frac{14\pi}{15}-\left(\frac{\pi}{2}\pm\arccos\left(\frac{5}{3}\sin\left(\frac{\pi}{15}\right)\right)\right)\)

Or:

\(\displaystyle \beta=\frac{13\pi}{30}\pm\arccos\left(\frac{5}{3}\sin\left(\frac{\pi}{15}\right)\right)\)

Using the Law of Sines again, we may state:

\(\displaystyle b=\frac{3\sin(\beta)}{\sin\left(\dfrac{\pi}{15}\right)}\)

And so the two possible values of \(b\) in km are:

\(\displaystyle b_1\approx2.07660968027428\)

\(\displaystyle b_2\approx7.70486632706377\)
 
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