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?

  • $\begingroup$ if h=hypotenuse,b=adjacent,p=opposite then slope of h=6:1 ,slope of b=1.5% calculate the length of b. $\endgroup$ – 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 $C$.

I suggest you go back to where you got this data and find the mistake there.

  • $\begingroup$ No. I didnt use pythagoras theorem here. nor did (4,3) influenced me. I was going by the steps. I used law of sines to find the sides a/sinA = b/sinB = c/sinC. I used (4,3) only to check if my line equation was right or not. Now I am unable to find the line equation of AC since I need slope. which is tan(some angle)! How do i find this angle? $\endgroup$ – Karthick Sep 10 '12 at 19:24
  • $\begingroup$ You found that angle in step 7, right? You say that value is wrong, but why do you think so? $\endgroup$ – Hew Wolff Sep 10 '12 at 19:31
  • $\begingroup$ See step 5: slope of BC = 0.58. This can also be calculated by m = y2-y1/x2-x1 = 10-3/16-4 = 0.58. This is where i am using the coordinates of C to check if the slope i found is right. It is right for BC. but not right for AC. The slope of AC would be = 15-3/10-4 = 2. I should get 2 as the slope, but i'm getting 1.43. ?!?!?!?!?! $\endgroup$ – Karthick Sep 10 '12 at 19:49
  • $\begingroup$ 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? $\endgroup$ – Hew Wolff Sep 10 '12 at 19:56
  • $\begingroup$ This is actually a question to find the third point (x,y) of a triangle. The book says "The third coordinate is (4,3)". I am unable to derive at it. There are five problems in the book. I've done the first two. I'm now stuck with this problem. For the first two questions, I found line equations of the two sides the triangles and then solved them to find their point of intersection. (4,3) is not any point on the line. It is the point of intersection C. $\endgroup$ – Karthick Sep 10 '12 at 21:46

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