# Creating specified triangle

1. May 5, 2006

### electronic engineer

how to create such a triangle in geometric way:

[AB]=5.5 cm, [AC]=3.7cm, ABC=100 (angle)

thanks

2. May 5, 2006

### Staff: Mentor

Not sure I understand the question, but check out the Law of Sines and the Law of Cosines. One of those might work for you....

3. May 6, 2006

### electronic engineer

no , what it's required is that to create with geometric way so not using trigonometric laws, i need to get that triangle by using only ruler and may be circuler

4. May 8, 2006

### Staff: Mentor

I'm not sure what a circuler is, but if you can use a ruler, then you just have to think about how to construct the 100 degree angle using the ruler. Why would knowing the arctan(10 degrees) help in this?

5. May 8, 2006

### Gokul43201

Staff Emeritus
ee : You must tell us what you have tried and where you are stuck. Please read the guidelines for posting here.

6. May 8, 2006

### daveb

Can you use both a protracter and a compass (or a compass that measures angles)?

7. May 9, 2006

### electronic engineer

yes,I can use them , you said both protractor and compass even though protractor itself is used to measure angles .... anyway to where are you going to get me?

8. May 9, 2006

### Gokul43201

Staff Emeritus
9. May 9, 2006

### VietDao29

Ok, I think I'll give you some hints to start off your problem:
First, given that AB = 5.5 cm
Can you construct the line segment AB?
From there, you know that: $$\widehat{ABC} = 100 ^ \circ$$. Can you find two rays, of which every point makes with AB an angle of 100o? You also know that AC = 3.7 cm. How can you find C?
Can you go from here? :)

10. May 9, 2006

### electronic engineer

yes i know that , i tried to follow this way before you tell me about it, the problem is how to find C because [AC]=3.7cm and the angle more than 90 (100 c) makes that length of 3.7cm not achievable especially that [AC]<[AB].....that's why i'm asking here , of course i don't want to get the others solve my problem , i'm just asking for guidance.

many thanks for you all

11. May 9, 2006

### Staff: Mentor

Well, I just sketched the triangle, and BC is about 7-something long. The triangle looks fine. So the problem is not that such a triangle doesn't exist, right EE? The problem is how to construct it with a ruler, protractor and compass, right?

So draw line AB from the origin to the right 5.5cm long. Then mark a point 3.7cm to the right of the origin and on the segment AB. Strike an arc with your compass so that it makes a circle of radius 3.7cm centered on the origin. So now point C has to lie on that line somewhere (actually, two somewheres), right? Now your only problem is how to construct a 100 degree angle. That's why I asked early on about the arctan(10 degrees)....

12. May 9, 2006

### GregA

Am I wrong to suggest that the question should have stated BC is 3.7 as opposed to AC is 3.7?...as VietDao suggests, draw two rays from B where the angle it makes with AB is 100 deg and you're gonna have difficulties making AC

Last edited: May 9, 2006
13. May 9, 2006

### Staff: Mentor

Don't you draw two rays from A? Although I guess I see what you mean, if "angle ABC" is supposted to have B as the common point of AB and BC. EE, are you sure it said "angle ABC = 100 degrees"? Or could it have meant angle CAB = 100 degrees?

14. May 10, 2006

### Gokul43201

Staff Emeritus
This is all you should have said in the beginning.

You are perfectly correct. The triangle with the given dimensions does not exist on a plane. There is an error in the question. Take it back to the teacher and ask them to correct it.