4.8 How to solve a triangle (scalene)

  Topography notes

To solve a scalene triangle we need three element: one of wich must be a side.

Triangle_with_notations_2

Now we explain all cases.

Case a) knowing one side and adjacent angles of a triangle: b, \alpha, \gamma

To solve and determine all other paramether of the triangle we follow these steps:

  1. We know: b, \alpha, \gamma, we looking for a,c,\beta
  2. The sum of internal angle into a triangle is a180°, then \beta=180^o-(\alpha +\gamma).
  3. Using the Law of sines we obtain other two side:
    a=\dfrac{b\sin{\alpha}}{\sin{\beta}}
    c=\dfrac{b \sin{\gamma}}{\sin{\beta}}
  4. Done.

Case b) Knowing two side and their angle of a triangle: a,b,\gamma

Steps:

  1. We know a,b,\gamma, we looking for c,\alpha, \beta
  2. Using Law of cosines we can determine c
     c= \sqrt{a^2 + b^2 -2ab \cos{\gamma}}
  3. similary we can obtain other angles solving Law of cosines by cosine:
    \alpha=arccos{\dfrac{b^2 + c^2 - a^2}{2bc}}
  4. and:
    \beta=arccos{\dfrac{c^2+a^2 - b^2}{2ca}}

There is another way to solve the scalene triangle in this case:

  1. Frist of all determine
    \alpha+\beta=180^o - \gamma ~~\rightarrow~~ \frac{\alpha+\beta}{2}=90^o - \frac{\gamma}{2}
    and
  2. \frac{\beta - \alpha}{2}=\arctan{ \left [ \dfrac{b-a}{b+a} \tan{(\frac{\beta + \alpha}{2})} \right ]}
    Now we have the parameters for solve triangle with:
  3. \beta = (\frac{\beta + \alpha}{2})+ (\frac{\beta - \alpha}{2})
  4. \alpha = (\frac{\beta + \alpha}{2})- (\frac{\beta - \alpha}{2})
  5. c=\dfrac{b \sin{\gamma}}{\sin{\beta}}~~=~~\dfrac{a \sin{\gamma}}{\sin{\alpha}}

Case c) knowing two sides and opposite angle of the triangle: a,b,\alpha

Follow these steps:

  1. By Law of sines determine another angle:  \beta=\arcsin{\left ( \dfrac{b \sin{\alpha}}{a}\right )}
  2.  \gamma = 180^o - (\alpha + \beta)
  3. and:
    c = \dfrac{a \sin{\gamma}}{\sin{\alpha}} ~~=~~\dfrac{b \sin{\gamma}}{\sin{\beta}}

Case d) Knowing all three sides: a,b,c

The faster way in a calculator is use law o cosines in inverse notation:

\alpha=\arccos{\dfrac{b^2 + c^2 - a^2}{2 b c}} \beta=\arccos{\dfrac{c^2 + a^2 - b^2}{2 a c}} \gamma=\arccos{\dfrac{a^2 + b^2 - c^2}{2 ab}}

There is another way with Briggs formulas:

  1. Determine semiperimeter: p=\frac{1}{2}(a+b+c)
  2. Use Briggs for determine tangents of half-angles:
\tan{(\dfrac{\alpha}{2})} = \sqrt{\dfrac{(p-b)(p-c)}{p(p-a)}} \tan{(\dfrac{\beta}{2})} = \sqrt{\dfrac{(p-c)(p-a)}{p(p-b)}} \tan{(\dfrac{\gamma}{2})} = \sqrt{\dfrac{(p-a)(p-b)}{p(p-c)}}

And then solve inverse functions for the value of angles.


Topography notes: Table of contentsby_nc_sa_2
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