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Triangles, what is the minimum amount of data required to...

by uperkurk
Tags: data, minimum, required, triangles
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uperkurk
#1
Mar1-13, 05:49 PM
P: 159
With triangles, what is the least amount of data I need before I can start working out other data about the triangle such as angles and lengths of sides ect?

I think it is at least 1 length and 2 angles or 2 lengths and 1 angle?

What about triangles other than right angle triangles?
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jbunniii
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Mar1-13, 06:14 PM
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If you know two angles, then that gives you the third because their sum must be ##\pi##. If this is all you know, then you can't determine the lengths of the sides (if you multiply all three sides by the same factor, then the angles stay the same), but you can determine their ratios.

If you know the lengths of all three sides, you can determine all three angles from this.

If you know the lengths of two sides and the angle between them, that is enough to determine the length of the third side and the other two angles.

If you know the length of one side and the angles at each end of that side, that is also enough to determine the rest.

All of the above is true whether or not it's a right triangle.
uperkurk
#3
Mar1-13, 07:13 PM
P: 159
Quote Quote by jbunniii View Post
If you know two angles, then that gives you the third because their sum must be ##\pi##. If this is all you know, then you can't determine the lengths of the sides (if you multiply all three sides by the same factor, then the angles stay the same), but you can determine their ratios.

If you know the lengths of all three sides, you can determine all three angles from this.

If you know the lengths of two sides and the angle between them, that is enough to determine the length of the third side and the other two angles.

If you know the length of one side and the angles at each end of that side, that is also enough to determine the rest.

All of the above is true whether or not it's a right triangle.
Thanks. I have some questions though that I just can't seem to work out. They give me 2 angles and 1 length. I can easily work out the 3rd angle but then that gives me 3 angles and 1 length.

But I don't know which rules to use in order to find the hyp, opp or adj.

jbunniii
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Mar1-13, 07:34 PM
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Triangles, what is the minimum amount of data required to...

Quote Quote by uperkurk View Post
Thanks. I have some questions though that I just can't seem to work out. They give me 2 angles and 1 length. I can easily work out the 3rd angle but then that gives me 3 angles and 1 length.

But I don't know which rules to use in order to find the hyp, opp or adj.
Can you post a specific example that is giving you trouble?
HallsofIvy
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Mar1-13, 07:39 PM
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jbunniii's list is based on the "congurence relations": SideSideSide, SideAngleSide, AngleSideAngle, ...

Yes if you have two angles you can solve for the third. So if you have two angles and any side you can, if one of the given angles was opposite the given side, you can find the third angle and apply "SideAngleSide".
uperkurk
#6
Mar1-13, 09:36 PM
P: 159
I don't know if I'm just being an idiot and missing the obvious but I'm trying all of the SohCahToa rules and none of them seem to work.

I'm checking my answers and it just isn't coming out with anything.
jbunniii
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Mar1-13, 10:10 PM
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The key thing about SohCahToa is that, unlike the rules I mentioned in my earlier post, SohCahToa only applies to right triangles.

In your figure, both triangles are right triangles, so SohCahToa applies.

In the triangle on the left, the hypotenuse is the side of length 22 cm. If we look at the angle labeled 40 degrees, the side opposite that angle is the one labeled X, and the side adjacent to the angle is the unlabeled side. Since we know the hypotenuse and want to find the opposite side, we want to use the "SOH" rule, i.e. ##\sin(40) = \textrm{opposite}/\textrm{hypotenuse} = x/22##, or equivalently, ##x = 22 \sin(40)##. Plugging this into a calculator (making sure it is set to "degrees"), I get ##x \approx 22(0.643) = 14.141## cm. How does that compare with your answer?
uperkurk
#8
Mar2-13, 01:00 AM
P: 159
Oh!!! Oh see, I was trying 22 DIVIDED by sin, cos tan 22. I see you did 22*sin40. Now I see.

So on the other triangle I see that the length marked x is adjacent to 18 so I use TOA? for which I get 18*tan55 = 25.7.

25.7^2+18^2 = 984.8 sqrt = 31.3 which would be the hyp.
jbunniii
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Mar2-13, 01:13 AM
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Quote Quote by uperkurk View Post
So on the other triangle I see that the length marked x is adjacent to 18 so I use TOA? for which I get 18*tan55 = 25.7.
Be careful to identify the opposite and adjacent sides correctly. The side opposite the 55 degree angle is 18, and the side adjacent to the angle is ##x##. Therefore, ##\tan(55) = \textrm{opposite}/\textrm{adjacent} = 18/x##, so ##x = 18/\tan(55)##.
uperkurk
#10
Mar2-13, 08:38 AM
P: 159
Quote Quote by jbunniii View Post
Be careful to identify the opposite and adjacent sides correctly. The side opposite the 55 degree angle is 18, and the side adjacent to the angle is ##x##. Therefore, ##\tan(55) = \textrm{opposite}/\textrm{adjacent} = 18/x##, so ##x = 18/\tan(55)##.
I'm confused now, in the first one you did 22*sin40. Now on the second one you did 18/tan55.
Mark44
#11
Mar2-13, 12:17 PM
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Quote Quote by uperkurk View Post
I'm confused now, in the first one you did 22*sin40. Now on the second one you did 18/tan55.
jbunniii is just doing ordinary algebra.

It would help if you wrote complete equations. In the first problem you cite, jbunniii started with sin(40) = x/22 and solved this equation for x.

In the second problem he started with tan(55) = 18/x, and solved this equation for x.
uperkurk
#12
Mar2-13, 02:48 PM
P: 159
I've gone back over what has been said and it makes much more sense now, I was just tired last night and wasn't thinking clearly.

The only bit that trips me up is when the only side given is the hyp, other than that I'm fine now thanks.
Vargo
#13
Mar4-13, 11:19 AM
P: 350
SAS, SSS, ASA, AAS, and SsA (meaning big side little side angle) are the definitive answers to your question. Consult Euclid Book I if memory serves. Given two angles, find the third using the fact that all three sum to 180. Otherwise, the Laws of Cosines and Sines are what you need to calculate the unknown data in every case.


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