Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am writing a program and I need to calculate the 3rd point of a triangle if the other two points, all sides and angles are known.

            A (6,14)
            ^
           / \
    14.14 /   \ 10.14
         /     \
        /       \
B (16,4)--------- C (x,y)
          10.98

A (6,14), B (16,4). Side AB is 14.14, AC is 10.14 and BC is 10.98 Angle A, B, C are 50, 45 and 85 degrees respectively...

I want to calculate the position of C. I don't know which formula to use for this. Actually i am performing triangluation. I am calculating the position of an incoming object (C).

share|improve this question
    
You can use parametric equations for the lines with those angles. Then, just take the intersection. But I guess that there is some formula with trigonometric informations. –  Sigur Aug 26 '12 at 14:23
    
Thanks! I've thought of this but as you said there got to be some formula for this! Got to pass high school again! –  Krat Aug 26 '12 at 14:30
    
@Sigur Yes you are right, trig is probably the way to go for this kind of problem. –  mathguy Aug 26 '12 at 15:28
    
Hint : One useful point will be, using distance formula. However it will be hard to use it but it will provide you at least one significant equation for(x,y). I think it would be $ x + y = -14.131 $ –  Rahul Taneja Aug 26 '12 at 18:05
    
@Krat : The singular is "vertex"; the plural is "vertices". I changed the title accordingly. –  Michael Hardy Aug 26 '12 at 19:08

3 Answers 3

It is enough to take the middle point of $BC$ and duplicate it. Let me explain:

Consider $M=1/2(AB+AC)$. Then $C=2BM$. Now you have the vector and its norm, so is enough to solve an equation.

share|improve this answer
    
I think if we find the equation of lines BC and AC, we can find their point of intersection. If we drop a perpendicular CD to AB, what will be the slope of AC? Is it CD/AD?? –  Krat Aug 28 '12 at 9:38
up vote 1 down vote accepted

Thanks everyone for the help! I found the answer. The formula for which is given here (Section: Intersection of two circles. It's not in the plot, but $d$ is the euclidean distance between the centers of the circles or $d=a+b$).

enter image description here

a = (r0 - r12 + d2 ) / (2d)

h = r0 sinP0 (or) r1 sinP1

P2 = P0 + a ( P1 - P0 ) / d i.e., x2 = x0 + a (x1 - x0) / d (and) y2 = y0 + a (y1 - y0) / d

x3 = x2 ± h ( y1 - y0 ) / d

y3 = y2 ± h ( x1 - x0 ) / d

share|improve this answer
    
I struggled for a while to understand where the formula for the coordinates of $P_2$ came from. The key is to realize that $(P_1 - P_0)/d$ is just the unit vector that points in the direction of $a$ (The original vector is $v = (P_{1y} - P_{0y}, P_{1x} - P_{0x})$), so multiplying it by $a$ gives the coordinates for $P_2$. –  nachocab Sep 7 at 15:59
    
And the key to the last step (the coordinates of $P_3$) is understanding that the unit vector along $h$ is perpendicular to the unit vector along $a$. –  nachocab Sep 7 at 20:29

Take the following figure I made in paint:
pic

So basically, you want to find the lengths $AD$ and $DC$. So lets make the coordinates $(X_A,Y_A),(X_B,Y_B),(X_C,Y_C)$. The length $AO = |Y_A - Y_B|$ and $BO = |X_A - X_B|$. That means that $ \tan(\angle ABO) = \frac{AO}{BO} \leftrightarrow \angle ABO = \arctan(\frac{AO}{BO})$.
Because the angles of a triangle sum up to $180$ degrees, $\angle BAO = 180 - \angle ABO - 90 \leftrightarrow \angle BAO = 90 - \angle ABO$.
$\angle DAC = \angle BAC - \angle BAO$.
$\sin(\angle DAC) = \frac{DC}{AC} \longleftrightarrow DC = AC \sin(\angle DAC)$
$\cos(\angle DAC) = \frac{AD}{AC} \longleftrightarrow AD = AC \cos(\angle DAC)$
$X_C = X_A + DC$
$Y_C = Y_A + AD$
Now if you want two formulas with all the information compacted, you get:
$X_C = X_A + AC\sin(\angle BAC - 90 + \arctan(\frac{AO}{BO}))$
$Y_C = Y_A + AC\cos(\angle BAC - 90 + \arctan(\frac{AO}{BO}))$
Just for the record, the angles and lengths are good information to have, but because of the way trigonometric functions work, the coordinates $(x,y)$ might appear on the opposite side of line $AB$ in your example, while still fulfilling your constraints. You may have to add some if statements if you want it going in a certain direction. Let me know if this equation works.

share|improve this answer
    
There got to be some easy way(s). One way as I have answered.. Another is to find the equations of lines AC and BC by using point-slope form y-y1 = m(x-x1). If CD is perpendicular to AB, we can find the angles BCD and ACD. From these angles can we find the slopes of BC and AC? –  Krat Aug 28 '12 at 9:52
    
@Krat CD and AB are not perpendicular (at least if you are referring to my image). Did you mean to say CD and AB? –  mathguy Aug 28 '12 at 10:07
    
not this image Sid! I am referring to the image I've drawn in the question.. –  Krat Aug 28 '12 at 10:19
    
@Krat What is point D on your picture? –  mathguy Aug 28 '12 at 22:37
    
D is a point on AB such that CD is perpendicular to AB. –  Krat Aug 29 '12 at 17:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.