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14 replies to this topic

#1Marques  Members

Posted 27 June 2001 - 07:26 PM

How can I find the angle of a triangle when only the length of each side are given? for example let say I have a triangle ABC and I''m trying to find angle B. Here are the dimensions of the triangle: Line AB = square root 2 or 1.4142135 AC = 1 BC = 1 I know just by looking at the triangle angle B is 45 degrees but how do I prove it using math. My first thought was to find the sine (opposite/hypotenuse) of angle B which give me the ratio of 1/1.4142135 or 0.7071068. If I look up sine 0.7071068 in a table this tell the angle B is a 45 degree angle. What I want to is know how to get the angle without using a look up table.

#2Anonymous Poster_Anonymous Poster_*  Guests

Posted 27 June 2001 - 07:44 PM

Use Pythagorem''s theorm.

Since line a is 1, line b is 1, then 1+1 = the square root of 2.

Since it is this theorm you can then say that angle a and angle b are 45 degrees each, thus giving you 90 degrees. 180 - 90 = 90 so angle b must be 90.

#3python_regious  Members

Posted 28 June 2001 - 04:13 AM

You don't actually have to use any trig or anything to prove that the triangle you are describing has angle B as 45 degrees.

Firstly, here is a picture of your triangle.

Now, because of pythag, angle C must be 90 degrees, otherwise AB cannot be root 2 in length. Because of this, angles A and B must both add to 90 degrees. Another thing about the triangle is that it is isoceles (spelling?), therefore angles A and B must be equal in size. So, from this, you can easily deduce that angles A and B MUST be 45 degrees ( 90/2 ).

Of course, to prove it in a more general sense, just use the cosine rule.

Edited by - python_regious on June 28, 2001 11:18:11 AM

#4Anonymous Poster_Anonymous Poster_*  Guests

Posted 28 June 2001 - 05:15 AM

You have to use law of cosines. Pythagorean thm will NOT work.
If you have corners A,B,C and sides a,b,c (a=length of side opposite angle A), then you get:
a^2=b^2+c^2-2*b*c*cos(A).
If a^2=b^2+c^2 you see that cos(A)=0 => A=pi/2, i.e. you get the converse of the pythagorean thm.

#5FloViel  Members

Posted 28 June 2001 - 05:40 AM

You can''t use Pythagorem''s theorm here because it''s only for the length of the sides and not for the angles. I would also suggest using the law of cosines.

#6flame_warrior  Members

Posted 28 June 2001 - 06:56 AM

Hi,
There is a much easier method.

First you know that two sides are equal. There is a rule that says if two sides are equal then the angles angles opposite to the sides are equal. Sum of the angles of triangles is 180 degrees. So the sum of the two angles is 90 because 180 - 90 = 90 degrees. Now the angles are equal so 90 / 2 = 45 degrees. This is a much simpler way and requires minimum calculations.

#7python_regious  Members

Posted 29 June 2001 - 12:00 AM

Thats what I tried to say.

#8flame_warrior  Members

Posted 29 June 2001 - 05:23 AM

Hi,

Well thats not pythogarem''s theorem. By the way if it wasn''t a right angled triangled the theorem will not be applicable.

The solution I provided is in general and can be applied to any triangle.

#9Anonymous Poster_Anonymous Poster_*  Guests

Posted 29 June 2001 - 08:09 AM

Thanks for your help, but a friend of my help me come up with a
solution to my problem. As you all know I was looking for a way
to find an angle when only the length of the side were known.
It turns out the solution was quite easy when someone show
upfront on a personal level. Unfortunely some guys didn''t
understand what I was looking for, but here it is for what its
worth.

First get the tangent ratio of the triangle but use the tan
inverse function on your calculator or arctangent(atan) which is
the same tangent inverse for C.
On your Calculator the calculation would be:
tan x^(-1) = opposite/adjacent
or in C:

So Let say we have triangle ABC
AB = hypotenuse
AC = 2 = opposite side
BC = 4 = adjacent

http://www.netcolony.com/entertainment/jmarques/images/inverse%20tangent.jpg

so there you have it
angle = atan(4/2);

angle = 26.565

#10Marques  Members

Posted 29 June 2001 - 08:20 AM

Thanks for your help, but a friend of mine helped me come up
with a solution to my problem. As you all know I was looking for
a way to find an angle when only the length of the sides are known.
It turns out the solution was quite easy when there
someone there to show you how. Unfortunely some you have
mis-interpet what I was saying. But I have a solution if there's
any who cares to know.

First get the tangent ratio of the triangle but use the tan
inverse function on your calculator or arctangent(atan) which is
the equivalent of tangent inverse for C.
On your Calculator the calculation would be:
tan x^(-1) = opposite/adjacent
or in C:

So Let say we have triangle ABC
AB = hypotenuse
AC = 2 = opposite side
BC = 4 = adjacent

so there you have it
angle = atan(4/2); // tangent inverse

angle = 26.565

Black Marq

btw you could probablely do the same with sine inverse and cosine
inverse I don't see why it wouldn't work. Tangent Inverse was the
only one my friend could remember doing.

Edited by - Black Marq on June 29, 2001 3:46:08 PM

#11Wizardry  Members

Posted 29 June 2001 - 08:30 AM

Keep in mind, however, that using the tangent only works on right triangles. It will not work for any other type of triangle. The problem with that is that you need to know the measure of one of the angles(the 90 degree one).

The law of cosines solution posted earlier is the correct one.

To review:

cos(A) = (b2 + c2 - a2)/(2bc)

Solving for A will give you the angle, no matter what type of triangle you''re dealing with.

Peace out.

#12Anonymous Poster_Anonymous Poster_*  Guests

Posted 29 June 2001 - 09:46 AM

quote:
Original post by Wizardry
Keep in mind, however, that using the tangent only works on right triangles. It will not work for any other type of triangle. The problem with that is that you need to know the measure of one of the angles(the 90 degree one).

The law of cosines solution posted earlier is the correct one.

To review:

cos(A) = (b2 + c2 - a2)/(2bc)

Solving for A will give you the angle, no matter what type of triangle you''re dealing with.

Peace out.

Thanks Wizardy I get what you saying and I apologize to the
others who are telling me about Law of Cosine for acting
like a stupid moron.

btw the way I have another stupid moron question in which its
guareentee to make your eyes roll into the back of you skull
Topic: Cosine Inverse
How are the ratios of cosine inverse converted back into angles?
In other words how does the cosine inverse really work.

#13BeerNutts  Members

Posted 29 June 2001 - 10:24 AM

Simple answer (Where Cos-1 = inverse cosine)
Cos(X) = Y
Cos-1(Cos(X)) = Cos-1(Y)
X = Cos-1(Y)

Basically, the inverse cosine of a cosine of an angle is the angle. The opposite is true also.

If you want a deeper answer, then I can answer with many more words.

Nutts

Edited by - BeerNutts on June 29, 2001 5:26:54 PM

#14Anonymous Poster_Anonymous Poster_*  Guests

Posted 29 June 2001 - 04:23 PM

quote:
Original post by BeerNutts
Simple answer (Where Cos-1 = inverse cosine)
Cos(X) = Y
Cos-1(Cos(X)) = Cos-1(Y)
X = Cos-1(Y)

Basically, the inverse cosine of a cosine of an angle is the angle. The opposite is true also.

If you want a deeper answer, then I can answer with many more words.

Nutts

Edited by - BeerNutts on June 29, 2001 5:26:54 PM

I not very good a trig not to mention inverse trig, but what are
X and Y used for?

#15Wizardry  Members

Posted 30 June 2001 - 05:25 AM

If you want a simple explanation, the inverse-cosine or arccosine is just the inverse of the cosine function like multiplication and division or addition and subtraction.

If you have the cosine of an angle, the inverse-cosine will give you the angle.

cos-1(cos(theta)) == theta

That''s about as complicated as it gets.

Peace out.

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