# Find the line equations of the sides of a triangle. (finding slopes)

                      A (10,15)
/\
/  \
/    \
/      \
/        \
C (x, y) /__________\ B (16, 10)


Given : $\angle BAC = 85^\circ$ ; $\angle ABC = 70^\circ$ ;

To find: Equations of $\overline{AC}$ and $\overline{BC}$ (or their slopes)

What I have done:

1. By distance formula I found $AB = \sqrt{61}$
2. By law of sines, I found the other values. $AC = 18.4103$ and $BC = 17.3661$
3. Found slope of $AB$. $m = -5/6$. Since slope is negative $m = -(-5/6) = 5/6$
4. $\arctan(5/6) = 39.80^\circ$. $\angle B$ is $70^\circ$. But it makes ($70^\circ - 39.80^\circ = 30.20^\circ$) with $x$-axis. $BC$ makes $30.2^\circ$ with $x$-axis.
5. Hence slope of $BC = \tan(30.2^\circ) = 0.5820$
6. $\Rightarrow (y - 10) = 0.582 (x - 16) \Rightarrow 0.582x - y = - 0.688$ (Equation of $\overline{BC}$)
7. $AC$ makes $55.2^\circ$ with $x$-axis. So slope is $\tan(55.2^\circ) = 1.4388$

The slope of $AC$ is false. The coordinates of $C$ is $(4,3)$. So if i cross check, the slope of $BC$ is correct but not of $AC$. Can anyone help me with this? How do i find the line equation of $\overline{AC}$? what will be its slope?

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if h=hypotenuse,b=adjacent,p=opposite then slope of h=6:1 ,slope of b=1.5% calculate the length of b. – user71404 Apr 7 '13 at 10:49

Your calculations are fine; there's a problem with your assumptions. For example, in step 2 you correctly find the lengths of AC and BC from the angles BAC and ABC. But if you find those lengths from the assumption $C = (4, 3)$ with the Pythagorean theorem, you'll get different values around 13. The angles are not consistent with that position for $C$.
It sounds like you constructed $C$ to be on the line $BC$ and now you're finding that it's not on the line $AC$. But that's not surprising; pretty much any point on the line $BC$ will not be on the line $AC$. The only point on $BC$ that will also be on $AC$ will be the actual position of $C$, and we don't know that yet. Does that make sense? – Hew Wolff Sep 10 '12 at 19:56