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                      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?

  • 0
    if h=hypotenuse,b=adjacent,p=opposite then slope of h=6:1 ,slope of b=1.5% calculate the length of b.2013-04-07

1 Answers 1