How to Create a Triangle with Given Dimensions and Angle?

[electronic engineer]

how to create such a triangle in geometric way:

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

thanks

 

[berkeman]

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...

 

[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

 

[berkeman]

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?

 

[Gokul43201]

how to create such a triangle in geometric way:

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

thanks

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

 

[daveb]

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

 

[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?

 

[Gokul43201]

... anyway to where are you going to get me?

To here : https://www.physicsforums.com/showthread.php?t=5374

 

[VietDao29]

how to create such a triangle in geometric way:

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

thanks

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 100^o? You also know that AC = 3.7 cm. How can you find C?
Can you go from here? :)

 

[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

 

[berkeman]

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)...

 

[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 going to have difficulties making AC

 

[berkeman]

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?

 

[Gokul43201]

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

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.