# Homework Help: Find lengths of triange sides using trig calculations

1. Mar 8, 2013

### pbonnie

1. The problem statement, all variables and given/known data
See attached image.

2. Relevant equations
Sine law
Cosine law

3. 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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2. Mar 8, 2013

### SteamKing

Staff Emeritus
Why don't you draw triangle ACD? Can you identify the lengths of any of the sides of this triangle?

3. Mar 8, 2013

### Ray Vickson

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.

4. Mar 8, 2013

### pbonnie

Oh okay, that makes sense. Except I don't have any angles, and all I have is one measurement. How would I proceed?

5. Mar 8, 2013

6. Mar 8, 2013

### Ray Vickson

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.

7. Mar 8, 2013

### pbonnie

The 90° angle?

8. Mar 8, 2013

Or DC is 1m?

9. Mar 8, 2013

### Ray Vickson

You need to figure out these things for yourself. We are not allowed to do your work for you.

10. Mar 8, 2013

### Staff: Mentor

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. Mar 8, 2013

### Ray Vickson

Sorry: I meant 8000m. Nevertheless, a 6000m high umbrella would still be a sight to see.

12. Mar 8, 2013

### SteamKing

Staff Emeritus
Is DC 1m long? Lay out the base and see if this is reasonable.

13. Mar 8, 2013

### pbonnie

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. Mar 8, 2013

### pbonnie

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. Mar 8, 2013

### Ray Vickson

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. Mar 8, 2013

### pbonnie

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. Mar 8, 2013

### Ray Vickson

My apologies: meant that BD is one-half a diagonal. Of course AD is given.

18. Mar 8, 2013

### pbonnie

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. Mar 8, 2013

### Ray Vickson

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.