# Any Suggestions

Started by Marques, Jun 27 2001 07:26 PM

14 replies to this topic

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#1
Members - Reputation: **210**

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.

Sponsor:

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#2
Anonymous Poster_Anonymous Poster_*
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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.

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.

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#3
Members - Reputation: **929**

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

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

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#4
Anonymous Poster_Anonymous Poster_*
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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.

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.

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#6
Members - Reputation: **122**

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.

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.

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#9
Anonymous Poster_Anonymous Poster_*
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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:

angle = atan(opposite/adjacent);

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

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:

angle = atan(opposite/adjacent);

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

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#10
Members - Reputation: **210**

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:

angle = atan(opposite/adjacent);

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

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:

angle = atan(opposite/adjacent);

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

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#11
Members - Reputation: **122**

Posted 29 June 2001 - 08:30 AM

Keep in mind, however, that using the tangent only works on

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

To review:

cos(A) = (b

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

Peace out.

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

To review:

cos(A) = (b

^{2}+ c^{2}- a^{2})/(2bc)Solving for A will give you the angle, no matter what type of triangle you''re dealing with.

Peace out.

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#12
Anonymous Poster_Anonymous Poster_*
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Posted 29 June 2001 - 09:46 AM

quote:Original post by Wizardry

Keep in mind, however, that using the tangent only works ontriangles. 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).right

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

To review:

cos(A) = (b^{2}+ c^{2}- a^{2})/(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.

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#13
Crossbones+ - Reputation: **3457**

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

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

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#14
Anonymous Poster_Anonymous Poster_*
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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?

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#15
Members - Reputation: **122**

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

That''s about as complicated as it gets.

Peace out.

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

cos

^{-1}(cos(theta)) == thetaThat''s about as complicated as it gets.

Peace out.