Any Suggestions
Author
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.
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.
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
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.
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.
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.
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.
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.
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
Author
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
This topic is closed to new replies.
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