Find lengths of triange sides using trig calculations

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Homework Help Overview

The discussion revolves around finding the lengths of triangle sides using trigonometric calculations, specifically applying the sine and cosine laws. The original poster expresses uncertainty about the relevance of certain measurements and angles in the context of a geometric problem involving a pyramid-like structure.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationships between triangle sides and angles, questioning the significance of the height and base measurements. There are attempts to identify relevant triangles and clarify the implications of given dimensions.

Discussion Status

Participants are actively engaging with the problem, offering suggestions to draw diagrams and reconsider measurements. Some guidance has been provided regarding the relationships between different triangle components, but no consensus has been reached on specific methods or solutions.

Contextual Notes

There are constraints regarding the available measurements, with some participants noting the lack of angles and the need for indirect measurements. The discussion also highlights the original poster's feelings of being unprepared due to a gap in recent math education.

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Homework Statement


See attached image.


Homework Equations


Sine law
Cosine law


The Attempt at a Solution


I'm really not sure what to do on this one. I know that the angle ABC and ACB will be the same, but after that I'm stumped. It's been a long time since I've taken a math course so I'm not sure what the relevance of the length of AD is.
 

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Why don't you draw triangle ACD? Can you identify the lengths of any of the sides of this triangle?
 
pbonnie said:

Homework Statement


See attached image.


Homework Equations


Sine law
Cosine law


The Attempt at a Solution


I'm really not sure what to do on this one. I know that the angle ABC and ACB will be the same, but after that I'm stumped. It's been a long time since I've taken a math course so I'm not sure what the relevance of the length of AD is.

AD tells you how tall the pyramid is. If you had AD = 0 it would be flat, with no peak at all. If you had AD = 6000m the thing would be almost as high as Mt. Everest.

You need to look at some other triangles, such as ADB.
 
Oh okay, that makes sense. Except I don't have any angles, and all I have is one measurement. How would I proceed?
 
Other than ADB (or ADC) which I think is 90°?
 
pbonnie said:
Oh okay, that makes sense. Except I don't have any angles, and all I have is one measurement. How would I proceed?

Actually, you have one measurement, and one "indirect" measurement---something you can easily figure out from the measurements you _do_ have. Just draw a picture, and look carefully at what you have.
 
The 90° angle?
 
Or DC is 1m?
 
pbonnie said:
Or DC is 1m?

You need to figure out these things for yourself. We are not allowed to do your work for you.
 
  • #10
Ray Vickson said:
AD tells you how tall the pyramid is. If you had AD = 0 it would be flat, with no peak at all. If you had AD = 6000m the thing would be almost as high as Mt. Everest.
Actually, 6000 m. short of the height of Mt. Everest by nearly 3000 m. The reference I found gives Everest's height as 29,035 ft or 8850 m. I remember the published height being 29,028 ft some years ago, but the Indian plate continues to push up the Himalayas, raising Everest in the process.
 
  • #11
Mark44 said:
Actually, 6000 m. short of the height of Mt. Everest by nearly 3000 m. The reference I found gives Everest's height as 29,035 ft or 8850 m. I remember the published height being 29,028 ft some years ago, but the Indian plate continues to push up the Himalayas, raising Everest in the process.

Sorry: I meant 8000m. Nevertheless, a 6000m high umbrella would still be a sight to see.
 
  • #12
Is DC 1m long? Lay out the base and see if this is reasonable.
 
  • #13
Ray Vickson said:
You need to figure out these things for yourself. We are not allowed to do your work for you.

I have no idea how to figure it out. If the lesson had properly explained it, then I wouldn't be asking for help. I do not have a teacher to ask these questions to so any help to actually help me understand so that I can then figure out the answer on my own is greatly appreciated.
 
  • #14
SteamKing said:
Is DC 1m long? Lay out the base and see if this is reasonable.

I'm not sure what you mean by lay out the base? I haven't taken any math courses in 5 years so I'm a little rusty. Thank you for your help
 
  • #15
pbonnie said:
I'm not sure what you mean by lay out the base? I haven't taken any math courses in 5 years so I'm a little rusty. Thank you for your help

The base is a square of sides 2m. The segment AD is one-half of a diagonal. Can you take it from there? That is, can you calculate the length of AD? If you can , then you will have a right triangle with known base and height, so you can calculate the hypotenuse; that will be the length of AB.
 
  • #16
I'm still not following. I know what AD is, it's 0.5m. I know that the triangle ACD is a right angle triangle. I know once I figure out DC then I can use Pythagorean theorem to find AC, which would be the same as AB. Then I can figure out the angles etc. The problem I'm having right now is figuring out DC. That is all I'm stuck on. Everything else is fine, I just need to know how to start.
 
  • #17
pbonnie said:
I'm still not following. I know what AD is, it's 0.5m. I know that the triangle ACD is a right angle triangle. I know once I figure out DC then I can use Pythagorean theorem to find AC, which would be the same as AB. Then I can figure out the angles etc. The problem I'm having right now is figuring out DC. That is all I'm stuck on. Everything else is fine, I just need to know how to start.

My apologies: meant that BD is one-half a diagonal. Of course AD is given.
 
  • #18
Oh! Does that mean, I can find BD by using Pyth. theorem and dividing that by 2? So c^2 = 2^2 + 2^2
...
so BD would equal 1.4 m??
I think I'm finally understand
 
  • #19
pbonnie said:
Oh! Does that mean, I can find BD by using Pyth. theorem and dividing that by 2? So c^2 = 2^2 + 2^2
...
so BD would equal 1.4 m??
I think I'm finally understand

BD is not equal to 1.4; it is equal to √2, and that is approximately 1.4 (but that is a pretty poor approximation). Don't bother using a decimal approximation at this stage; just keep the length as √2 .

Note: that type of thing can come back to bite you later; it is important to NOT use such approximations except where you need them, especially in larger, more complicated problems.
 

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