Angle Measures of a Triangle Question

whitley

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The degree measures of a triangle are x, y and z and all are integers.

Angles measures must be x<y<z

What is the largest value of z?
What is the smallest value of z?
Can y be twice x and z be twice y?
How about 4 times each?
Any multiple?

All anles must be integers
 
What is the angle-sum of any triangle?

Working within this restriction, and keeping in mind that x can't be zero, x is strictly smaller than y, and y is strictly smaller than z, what is the largest value that z can have?

What is the smallest value that z can have?

If y = 2x and z = 2y and you have the given angle sum, can you find a solution?

If y = 4x and z = 4y, can you find a solution?

If y = mx and z = my, can you find a solution?

What have you tried? How far have you gotten?

Eliz.
 
whitley said:
The degree measures of a triangle are x, y and z and all are integers.
Angles measures must be x<y<z
What is the largest value of z?
What is the smallest value of z?
Can y be twice x and z be twice y?
How about 4 times each?
Any multiple?
All anles must be integers
What do you know about triangles and degrees?
How many degrees in a triangle?
If smallest angle = 40 degrees and there is a difference
of 10 degrees in the other 2 angles, what size are these 2 angles?

Not picking on you...but not interested in a whole lot of typing
covering stuff you may already know...
 
I know the sum the of angles of a triangle = 180.

I think the biggest angle measure z can be is 177. Then y would be 2 and x would be 1, but Iam having trouble with formula to determine this.

I also think the smallest z could be is 89. Then y would be 46 and x would be 45, but once again the what is the formula to determine this.


I think that there cannot be multiples of each other.

Help!!!
 
Could there be a 59° - 60° - 61° triangle?

If y=2x & z=2y then x+2x+4x=180. Is there a solution?
 
whitley said:
I know the sum the of angles of a triangle = 180.
Correct.

whitley said:
I think the biggest angle measure z can be is 177. Then y would be 2 and x would be 1, but Iam having trouble with formula to determine this.
To a certain extent, you'll have to use common sense and argumentation, rather than an "equation", because your answer is restricted to being whole numbers. ("Integers" is a bit misleading: you can't have a triangle with an angle measure that's zero or negative.)

I agree with your answer.

whitley said:
I also think the smallest z could be is 89. Then y would be 46 and x would be 45, but once again the what is the formula to determine this.
Can't you get closer than that? (Hint: Divide 180 by 3. Then fiddle with that result.)

whitley said:
I think that there cannot be multiples of each other.
Why not? What have you tried?

Eliz.
 
Is there an equation that can be used to do this?

I agree with the 61, 60 and 59.
 
Is there an equation to prove that the angles are 177, 2 , and 1.
Is there an equation to prove that the angles are 61, 60, and 59.

Is there an equation to prove that angles can y=2x and z=2y?
Is there an equation to prove that angles can be y=4x and z=4y?

Or is there an equation for any multiple?
 
I am sure that this will sound picky to you.
But no equation has ever proved anything.
In this case, as in most cases, logic proves anything that can be proved.
Can you logically defend your answers?
If so then you have a proof.
 
As for "equations" for the "multiples" cases, you have been provided the set-up. Are you going to pursue the solutions at some point? (There is no guarantee that somebody will be posting the full solution for you. You might want to get started working on this yourself.)

Eliz.
 
Since y > x, then we can say: y = x + a
Since z > y, then we can say: z = x + a + b

Then we have:
x + x+a + x+a+b = 180
3x + 2a + b = 180
3x = 180 - 2a - b
x = (180 - 2a - b) / 3 ... x = 180/3 - (2a + b)/3

Since 180 is divisible by 3, then 2a+b must also be divisible by 3;
so 2a+b = 3, 6, 9, ... , 171, 174, 177.
if 177, then x = 1 (minimum)
if 3, then x = 59 (maximum)

That's just an example of how to "experiment" ...
 
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