Finding a point on a circle/triangle
- Thread starter TPiddy
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It might help if you shared that information (since the solution method may very depend upon the information available). :idea:TPiddy said:
What are the coordinates of A and B? What are the lengths of the three sides? What are the measures of the three angles? :?:
When you reply, please show everything you've tried so far. Thank you!
Eliz.
:arrow: P.S. Welcome to FreeMathhelp!
This is actually a dynamic application that I am using this for... so the point values will vary based on the users mouse movement.
The center is a fixed point (0, 0) and is angle B in this application. The top is a fixed point, (-5, -51) and the user's mouse is always going to create an angle at B that is less than 180 degrees.
AB and BC are the sides that have the same length.
The center is a fixed point (0, 0) and is angle B in this application. The top is a fixed point, (-5, -51) and the user's mouse is always going to create an angle at B that is less than 180 degrees.
AB and BC are the sides that have the same length.
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Um... so this would appear actually to be a much-more involved exercise than it originally seemed. :shock:TPiddy said:
Earlier, B was given as being a corner of the triangle. How is B now an "angle" but also a fixed point? Of what is B the "center"?TPiddy said:
Since the "center", (0, 0), is higher than (-5, -51), in what sense is the fixed point the "top"? Of what is it the top?TPiddy said:
Please reply with the full text of the exercise. Thank you!
Eliz.
Well, this is not an exercise, it's something I created...
B is a fixed point... but I suppose I should have said that the angle I know is ABC...
Also, this is a Flash application, so the Y coordinates run top to bottom rather than bottom to top.
So I know the following:
Point A (0, -45) (I moved it);
Point B (0, 0);
AB (45)
BC (45)
Angle ABC (130) - This angle will vary so I need a formula that I can create that will give me the accurate results of Point C regardless of what the angle is...
So, it's an isosceles triangle with point B being the middle point where the two sides that are the same length meet.
B is a fixed point... but I suppose I should have said that the angle I know is ABC...
Also, this is a Flash application, so the Y coordinates run top to bottom rather than bottom to top.
So I know the following:
Point A (0, -45) (I moved it);
Point B (0, 0);
AB (45)
BC (45)
Angle ABC (130) - This angle will vary so I need a formula that I can create that will give me the accurate results of Point C regardless of what the angle is...
So, it's an isosceles triangle with point B being the middle point where the two sides that are the same length meet.
pka
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I can give you a general solution. But I would not want to do the algebra required to solve it.
For the purpose of notation: \(\displaystyle a = \left( {a_1 ,a_2 } \right),\;b = \left( {b_1 ,b_2 } \right),\;c = \left( {x ,y } \right),\;\theta = \angle ABC\).
Note that if we know any angle is an isosceles, then we know all angles.
Because AB=BC: \(\displaystyle \left( {a_1 - b_1 } \right)^2 + \left( {a_2 - b_2 } \right)^2 = \left( {x - b_1 } \right)^2 + \left( {y - b_1 } \right)^2\).
Using vector geometry: \(\displaystyle \cos \left( \theta \right) = \frac{{\left( {a_1 - b_1 } \right)\left( {x - b_1 } \right) + \left( {a_2 - b_2 } \right)\left( {y - b_2 } \right)}}{{\sqrt {\left( {a_1 - b_1 } \right)^2 + \left( {a_2 - b_2 } \right)^2 } \sqrt {\left( {x - b_1 } \right)^2 + \left( {y - b_2 } \right)^2 } }}\).
Now you have two equations in two unknowns. It is somewhat easier if B=(0,0).
For the purpose of notation: \(\displaystyle a = \left( {a_1 ,a_2 } \right),\;b = \left( {b_1 ,b_2 } \right),\;c = \left( {x ,y } \right),\;\theta = \angle ABC\).
Note that if we know any angle is an isosceles, then we know all angles.
Because AB=BC: \(\displaystyle \left( {a_1 - b_1 } \right)^2 + \left( {a_2 - b_2 } \right)^2 = \left( {x - b_1 } \right)^2 + \left( {y - b_1 } \right)^2\).
Using vector geometry: \(\displaystyle \cos \left( \theta \right) = \frac{{\left( {a_1 - b_1 } \right)\left( {x - b_1 } \right) + \left( {a_2 - b_2 } \right)\left( {y - b_2 } \right)}}{{\sqrt {\left( {a_1 - b_1 } \right)^2 + \left( {a_2 - b_2 } \right)^2 } \sqrt {\left( {x - b_1 } \right)^2 + \left( {y - b_2 } \right)^2 } }}\).
Now you have two equations in two unknowns. It is somewhat easier if B=(0,0).
Wow.... I don't understand that at all.... what I was thinking was more of trying to find them individually.... something like adding a second triangle at a 90 degree angle, that way I can take the x from one point of it and the y from another....
Nice diagram...even if A is (0,-45) :shock:
Easy to calculate b: 2*45*COS(20) ; ~84.57
Extend AB to D, such that ADC is a right triangle;
easy to calculate BD and CD; these will be the coordinates;
but how you translate them from (0, -45) I'll leave for pka :idea:
Easy to calculate b: 2*45*COS(20) ; ~84.57
Extend AB to D, such that ADC is a right triangle;
easy to calculate BD and CD; these will be the coordinates;
but how you translate them from (0, -45) I'll leave for pka :idea:
TP, if you can't "see" that if 90 dead on, then coordinates are (45,0),TPiddy said:
then I wonder why you're tackling this.
Anyhoo, now that I got suckered into this:
to start, there is no need to calculate b (or AC);
simply create right triangle BCD (right angle at D of course);
angle CBD becomes 40 degrees (twice 20 degrees); so angle BCD = 50 degrees.
So you have a right triangle with hypotenuse = 45;
so CD = 45 * SIN(40) = ~28.93
and BD=45 * COS(40) = ~34.47
So C's coordinates = (28.93, 34.47)
For general case, let r = radius and u = given angle (140 in your example);
then the coordinates (x,y) are:
x = r * SIN(180-u)
y = r * COS(180-u)
And YES, it will work is u is less than 90;
the y coordinate will be negative, of course.
If you want to really SEE it work, get some graph paper and draw to scale
your r=45 and u=140 example, including right triangle BDC:
then rotate sheet a half turn: you'll see why the coordinates are both positive.
For u < 90 example, try u=70:
you'll get x = ~42.29 and y = ~-15.39;
again, include right triangle BDC (BD will be along x-axis);
rotate half turn and you'll SEE why y is negative.
That's it; I suggest you get familiar with Law of Sines and Cosines.
Ok, I've got something similar established here... but I'm getting negative results for some reason....
For example... if my angle is 129, the angle used in the sine is 51 (180 - 129), the cosine is approximately -0.69, forcing my y coordinate into the negatives even though it should clearly be a positive number.... Am I doing something wrong here?
Hmm... hold on... I don't think I'm getting accurate results from Flash's math compiler.
For example... if my angle is 129, the angle used in the sine is 51 (180 - 129), the cosine is approximately -0.69, forcing my y coordinate into the negatives even though it should clearly be a positive number.... Am I doing something wrong here?
Hmm... hold on... I don't think I'm getting accurate results from Flash's math compiler.