Trig Word Proble

[shooterman]

There is a women on top of a light house that is 2800 ft tall and she sees a yacht and a barge in the distance.
She measured the angle of depression of the yacht to be 20 degrees and the angle of depression of the barge to be
15 degrees and 30 min. She thinks the yacht and the barge to be 300 ft apart. If they are less than 300 ft apart than she would sound the horn. Are they closer than 300ft? and will she sound the horn.

So i drew the problem to make it visual.
[attachment=0sx6kpk0]th_Problem.jpg[/attachmentsx6kpk0]

To figure how far the yacht and barge are apart i solved for x which is the distance from light house for each triangle with this equation.
Then subtract the distance of the big triangle from small.

sin 20 Degrees = 2800/x for the small triangle and for the big triangle sin 15 degrees 30'= 2800/x
Am i puting it in the correct equation?

I got 58 ft as an answer.

 

[mmm4444bot]

It would be very dangerous for a person to stand on the top of a lighthouse; people generally stand on the platform built for this purpose.

So, let's assume that the lighthouse is NOT 2,800 feet tall. Let's assume instead that the women's eyeballs are 2,800 feet above the surface of the water.

You've used the same variable to represent two different numbers. That's not a good idea. Once you pick the symbol x to represent a length, you must pick different symbols to represent different lengths.

How about the following approach, instead?

You have two right-triangles.

You know all three interior angles in each of these two right-triangles.

You know the height of each.

Use the Law of Sines to determine the base of each.

Subtract the shorter base from the longer base.

Does this make sense ?

If not, then please ask specific questions.

 

[Deleted member 4993]

There is a women on top of a light house that is 2800 ft tall and she sees a yacht and a barge in the distance.
She measured the angle of depression of the yacht to be 20 degrees and the angle of depression of the barge to be
15 degrees and 30 min. She thinks the yacht and the barge to be 300 ft apart. If they are less than 300 ft apart than she would sound the horn. Are they closer than 300ft? and will she sound the horn.

So i drew the problem to make it visual.
[attachment=0:3vhb8vo5]th_Problem.jpg[/attachment:3vhb8vo5]

To figure how far the yacht and barge are apart i solved for x which is the distance from light house for each triangle with this equation.
Then subtract the distance of the big triangle from small.

sin 20 Degrees = 2800/x for the small triangle and for the big triangle sin 15 degrees 30'= 2800/x
Am i puting it in the correct equation?

I got 58 ft as an answer.

The distances are proportional to the cotangent of those angles of depression.

so

the distance = 2800 * [cot (22°) - cot(15.5°)]

 

[shooterman]

"The distances are proportional to the cotangent of those angles of depression.

so

the distance = 2800 * [cot (22°) - cot(15.5°)]"

=3027.92
Distance of what?

 

[shooterman]

Oh wait its wrong.

 

[shooterman]

= -2402.24
Distance of What

 

[mmm4444bot]

= -2402.24

Distance of What

2800*[cot(22°) - cot(15.5°)] = -3166, rounded to a whole number.

Using the approach that I previously posted, I get a distance of roughly 3,399 feet between the yacht and the barge.

Cheers

 

[Denis]

Shooter, WHY did you post 2800 as tower's height? Obviously it's 280.
Ever see a tower over half a mile high? :shock:

 

[mmm4444bot]

… Obviously it's 280…

The assumption that some person is standing on top of a 2,800' lighthouse is exactly the motivation for my posted sarcasm above.

However, the senario could be a 5'5'' woman standing on a 95' outlook platform built 2,700' above sea level. :idea:

 

[Denis]

Re:

[However, the senario could be a 5'5'' woman standing on a 95' outlook platform built 2,700' above sea level. :idea:
[/color]

you mean on a 94'5" platform, right?!

Shooter has always been careless in the past with his postings...looks like he's carrying on...

 

[mmm4444bot]

you mean on a 94'5" platform, right?! No. I mean her eyeballs are 5 feet above the platform. ANYTHING, really, that's plausible!

Shooter has always been careless in the past with his postings...looks like he's carrying on... Yup. Also, got stumped by the number 3/3, if memory serves.