Finding the Distance Between Buoys: A Cruise Ship Balcony Problem

[mathdrama]

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) .

 

[MarkFL]

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 $d_2$.

 

[mathdrama]

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?

 

[MarkFL]

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:

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

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

Now use the Law of Cosines:

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

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

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.

 

[CellCoree]

I would approach this problem by first drawing a diagram to visualize the situation. From the given information, we know that the person on the balcony is looking down at the two buoys, and that the first buoy is directly east at an angle of depression of 32°, while the second buoy is 65° south of east at an angle of depression of 40°.

Using basic trigonometry principles, we can determine that the angle between the two buoys, B1 and B2, is 65° - 32° = 33°. We can also use the tangent function to find the distance between the buoys, x, by setting up the following equation:

tan(33°) = (25m / x)

Solving for x, we get x = 25m / tan(33°) ≈ 41.11m. Therefore, the distance between the two buoys is approximately 41.11 meters.

It is also important to note that this calculation assumes that the buoys are at the same level as the person on the balcony, and that the sea level is flat. If the buoys are at different depths or the sea level is not flat, the calculation would be more complex and require additional information.