4.8 How to solve a triangle (scalene)

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

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: $a,b,\alpha$

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.