Finding the Distance Between Buoys: A Cruise Ship Balcony Problem

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Discussion Overview

The discussion revolves around a geometry problem involving the calculation of the distance between two buoys as observed from a cruise ship balcony. The problem includes angles of depression and requires the application of trigonometric principles, specifically the Law of Sines and the Law of Cosines, as well as the tangent function.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses uncertainty about how to approach the problem and the placement of angles.
  • Another participant suggests drawing a sketch and proposes using the Law of Sines and Law of Cosines to find the distances.
  • A later reply emphasizes the importance of using exact values in calculations rather than rounded values to avoid compounding errors.
  • Specific formulas for calculating distances to the buoys are presented, including the use of cotangent functions and the Law of Cosines.
  • There is a discussion about the convenience of posting work versus attaching files, with a preference for posting work for better visibility.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the best approach to solve the problem, and there are differing opinions on the use of rounded versus exact values in calculations. The discussion remains unresolved regarding the optimal method for finding the distance between the buoys.

Contextual Notes

Participants have not explicitly stated all assumptions, and there may be dependencies on the definitions of angles and trigonometric functions used in the calculations. The discussion also reflects varying levels of understanding regarding the application of trigonometric principles.

mathdrama
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Not really sure how to do this problem. I'm not even sure where the angles are.

5. The balcony of a cruise ship is 25m above sea level. A person standing on the balcony sees two buoy’s in the water below. The first buoy is situated directly east of her at an angle of depression of 32°. The second buoy is situated 65° south of east at an angle of depression of 40°. Find the distance (x) between the two buoys (B1 and B2) .
 

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I would begin by drawing a sketch:

View attachment 2514

Now, you can use the Law of Sines to find $d_1$ and $d_2$, and then the Law of Cosines to find $d$, the distance between the buoys.

edit: You could also consider using the tangent function to find $d_1$ and $d_2$.
 

Attachments

  • buoys.jpg
    buoys.jpg
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MarkFL said:
I would begin by drawing a sketch:

View attachment 2514

Now, you can use the Law of Sines to find $d_1$ and $d_2$, and then the Law of Cosines to find $d$, the distance between the buoys.

edit: You could also consider using the tangent function to find $d_1$ and $d_2$.

Are attachments acceptable or would LaTex still be more convenient?

Could you kindly help me check my work?
 

Attachments

Posting your work here rather than attaching it is much more convenient for those looking at your work.

One issue I have with what you have done, and it is certainly something I have seen many students do, is using rounded values in your computations. It is better to use exact values until the very end, and only then use a decimal approximation is so desired. Errors from rounding can become compounded if used in intermediary steps.

This is how I would work the problem:

$$d_1=25\cot\left(32^{\circ}\right)$$

$$d_2=25\cot\left(40^{\circ}\right)$$

Now use the Law of Cosines:

$$d^2=25^2\cot^2\left(32^{\circ}\right)+25^2\cot^2\left(40^{\circ}\right)-2\cdot25^2\cot\left(32^{\circ}\right)\cot\left(40^{\circ}\right)\cos\left(65^{\circ}\right)$$

$$d=25\sqrt{\cot^2\left(32^{\circ}\right)+\cot^2\left(40^{\circ}\right)-2\cot\left(32^{\circ}\right)\cot\left(40^{\circ}\right)\cos\left(65^{\circ}\right)}\approx38.4813948\text{ m}$$

You see, using rounded values caused you to round up when the true value should be rounded down. Other than this issue though, your method was correct.
 

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