Solving a triangle without a known value

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Duckss2211

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Hello forum :) My son and I were working through some test practice questions last night, and while I though I had a pretty good handle on the general concept of geometry, trig and algebra, this one had me stumped as to where to begin...

A triangle has internal angles of 120deg, 40deg and 20deg. The shortest side is 10mm shorter than the longest side. What are the lengths of all the sides?

If someone can point me in the right direction it would be much appreciated.

Cheers.
 
Duckss2211 said:
Hello forum :) My son and I were working through some test practice questions last night, and while I though I had a pretty good handle on the general concept of geometry, trig and algebra, this one had me stumped as to where to begin...

A triangle has internal angles of 120deg, 40deg and 20deg. The shortest side is 10mm shorter than the longest side. What are the lengths of all the sides? If someone can point me in the right direction it would be much appreciated. Cheers.
Click to expand...
As with any geometry problem - first draw a sketch.

Then tell us what are the laws of triangles that you know - involving lengths of the sides and measures of internal angles.
 
I have drawn the sketch, and as I mentioned, am quite familiar with the principals of Trig when it comes to right angle triangles, not so much with oblique triangles. It's the "+10mm" bit that has thrown me.
 
Duckss2211 said:
My son and I were working through some test practice questions last night, and while I though I had a pretty good handle on the general concept of geometry, trig and algebra, this one had me stumped as to where to begin...
A triangle has internal angles of 120deg, 40deg and 20deg. The shortest side is 10mm shorter than the longest side. What are the lengths of all the sides?
Click to expand...
I would use numbers for the angle measures: \(\frac{2\pi}{3},~\frac{2\pi}{9}~\&~\frac{\pi}{9}\)
In a triangle the shortest side is opposite the least angle; as well as the longest side is opposite largest angle.
 
Thanks pka, but I'm sorry, I can't see how that helps?
So that makes the angles out to 0.349, 0.698 & 2.094 as portions of pi?
 
Duckss2211 said:
I have drawn the sketch, and as I mentioned, am quite familiar with the principals of Trig when it comes to right angle triangles, not so much with oblique triangles. It's the "+10mm" bit that has thrown me.
Click to expand...
Can you post an image of the sketch including the vertices name - A, B & C.?
 
Duckss2211 said:
A triangle has internal angles of 120deg, 40deg and 20deg. The shortest side is 10mm shorter than the longest side. What are the lengths of all the sides?
Click to expand...
I presume you have sketched the triangle, putting x and x-10 on the appropriate sides.

Then how about using the Law of Sines?

EDIT: I see you just posted a picture. There are two problems: You only need one variable; and you didn't put A-10 on the right side.
 
Duckss2211 said:
View attachment 23039
Click to expand...
Duckss2211 said:
A triangle has internal angles of 120deg, 40deg and 20deg. The shortest side is 10mm shorter than the longest side. What are the lengths of all the sides?
Click to expand...
According to your drawing:

longest side = AC = x

shortest side = AB = x - 10

What are the triangle laws that your son knows that will give relations between the sides (AB , BC & CA) and measure of angles (ABC, BCA and CAB)?
 
Duckss2211 said:
C = A-10 and therefore A = C+10
Correct?
Click to expand...
No.

It is not your "C" that equals A-10. That isn't the shortest side!

Also, stating both C = A-10 and A = C+10 doesn't help, since those are equivalent. Try doing what I suggested, and use one variable, whether x or A. Then use the Law of Sines, unless you haven't learned it.

Please be aware, too, that your labeling is nonstandard (though perhaps it is what you were taught). We usually name points A, B, and C, and the opposite sides a, b, c. Your labels are very confusing, though that doesn't make it impossible to do the work.
 
Thank you Dr.Peterson - that's what you get for doing this stuff on the side while busy at work... lol.
My bad... I sketched it and never revisited which values I put where.
I am 50 years old and much of what I know and how I label things aren't from classical teachings, more life experiences. I was just trying to help my lad get his head around the algebra/trig combination. I can appreciate that people are trying to walk me through to some kind of epiphany, but that's not working... hahahah
 
I presume your son is checking what you do against what he has learned! Hopefully he can see how to apply the Law of Sines once you get the picture right.
 
A couple of points.

There are several ways to measure angles. The way that is most useful in calculus is measurement by radians. The common way in the US is degrees, and 90% of the population never works with radians. In post # 4, the measurements of the angles were given in radians. There is absolutely no need to use radians; degrees will work just fine.

It is possible to solve this problem using what you know about right triangles, but it is far easier to use the Law of Sines, which is explained under that title at Wikipedia.

If you decide to go the right angle approach (probably because your son is expected to solve it without the Law of Sines), I personally would disregard Dr. Peterson’s advice to use a single variable, but I would follow his advice to label the sides in standard form, a is the length of the side opposite point A, etc. That gives you three unknowns, a, b, and c. Letting a be the length of the longest side, construct a perpendicular to A from the line segment BC. Name as D the point where that perpendicular intersects BC. You now have two right triangles to work with, ABD and ACD. You also have three additional unknowns, x the length of BD, y the length of CD, and z the length of AD. Six unknowns requires six equations. Two should be obvious. The rest require a bit of basic trig.

As a starting point, draw a sketch as I suggested and figure out the angles in the two right triangles.
 
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