# Find point given a line and two angles

Let's say I have two points $p_1=(x_1, y_1)$ and $p_2=(x_2, y_2)$, which are given as two points of a triangle $T$. And let's say I know the angles of $T$ at $p_1$ and $p_2$. How do I find the third point $p_3=(x_3, y_3)$ of $T$?

• "Find point given a line and two *angles" – Argon May 15 '12 at 1:14
• Can't you ask the two angels? :) – user17762 May 15 '12 at 1:14
• @Vakey Do you mean you know the angles subtended by $(x_3,y_3)$,$(x_1,y_1)$ at $(x_2,y_2)$ and $(x_3,y_3)$,$(x_2,y_2)$ at $(x_1,y_1)$? – user17762 May 15 '12 at 1:31
• Have a look at any exposition on "angle side angle"; for example mathsisfun.com/algebra/trig-solving-asa-triangles.html – Fixee May 15 '12 at 2:42
• @Vakey Have a look at zonalandeducation.com/mmts/intersections/… – Fixee May 17 '12 at 16:14

## 1 Answer

I believe that this is what you want. We can use basic linear algebra.

We have the two sides of the triangle. I write these sides as lines:

$$f(x)=m_1x+b_1=\frac{y_1-y_3}{x_1-x_3}x+b_1$$ $$g(x)=m_2x+b_2=\frac{y_2-y_3}{x_2-x_3}x+b_2$$

$m$ is the slope of the line. I assume that the angles you are talking about are the angles of elevation or depression of the lines $f$ and $g$. To turn an angle into slope, we use $\tan \theta$ where $\theta$ is the angle. Thus

$$f(x)=\tan (\theta_1)x+b_1$$ $$g(x)=\tan (\theta_2)x+b_2$$

To find $b_1$ and $b_2$, solve for them in $f(x_1)=y_1$ and $g(x_2)=y_2$ respectively.

$$f(x_1)=y_1=\tan (\theta_1)x_1+b_1\implies b_1=y_1-\tan (\theta_1)x_1$$ $$g(x_2)=y_2=\tan (\theta_2)x_2+b_2\implies b_2=y_2-\tan (\theta_2)x_2$$

To find $x_3$, we know this is the unique point such that $f(x_3)=g(x_3)$. Thus

$$f(x_3)=g(x_3)=\tan (\theta_1)x_3+y_1-\tan (\theta_1)x_1= \tan (\theta_2)x_3+y_2-\tan (\theta_2)x_2 \implies x_3= \frac{\tan (\theta_1)x_1-\tan (\theta_2)x_2+y_2-y_1}{\tan (\theta_1)-\tan (\theta_2)}$$

I'm sure there are other ways to do this, but this is the first thing that comes to mind.

• Can you please check line functions . They should be $f(x)=\tan (\theta_1)x+b_1$ and $g(x)=\tan (\theta_2)x+b_2$. should not? – Mathlover May 15 '12 at 12:53
• Thanks. Though I am having trouble finding x3. I used an example where x1=0, y1=4, x2=4, y2=0, θ1 = 45, θ2 = 45. Pointing this out on a graph, I have found that x3=4 and y3=4. Using your formula above I have found that arctan(θ1)−arctan(θ2) = arctan(45)−arctan(45) = 1.5-1.5 = 0. And you can't divide by zero. Is my math wrong, or is this a special case? – Vakey May 15 '12 at 13:27
• @Mathlover Indeed it is $\tan$, my mistake. I never remember which one to use! – Argon May 15 '12 at 19:48
• @Vakey They are not both $45^\circ$. One should be $-45 ^\circ=315 ^\circ$ – Argon May 15 '12 at 19:55
• @Argon Thanks! That works. However, how do you get y3? – Vakey May 16 '12 at 12:59