Let say you have a triangle
Angle A = 41 degrees , side b = 3.41 and c = 5.83
can you use pythagoras theorem to find the side a?
and how can you find Angle B and C
[Math] Solving a triangle, given two sides and the measure of the included angle
algebra-precalculusgeometrytrianglestrigonometry
Related Solutions
While the other answers have covered the basic use of the Law of Sines, they've all missed a critical point: there are two triangles that fit your given information. Here is the triangle you've found:
But, when you're solve $\sin B=\frac{b\sin A}{a}$, there are two solutions that could be angles in a triangle, $B_1=\arcsin\left(\frac{b\sin A}{a}\right)$ (the one you found) and $B_2=180°-\arcsin\left(\frac{b\sin A}{a}\right)$. Using this second possible measure for $B$ won't always yield a triangle, but in this case it does:
Here are both triangles pictured together:
This issue with the Law of Sines, sometimes called the "ambiguous case," is almost certainly what the picture in your question here is about.
See also my answer here on general techniques for triangle-solving.
Consider the following triangle $\triangle ABC$, the $ \color{maroon} {\text{poly 1}}$ below, with sides $\color{maroon}{\overline{AB}=c}$ and $\color{maroon}{\overline{AC}=b}$ known. Further the angle between them, $\color{green}\alpha$ is known.
$\hskip{2 in}$
Then, the law of cosines tell you that $$\color{maroon}{a^2=b^2+c^2-2bc\;\cos }\color{blue}{\alpha}$$
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Best Answer
There isn't a simple way to use the Pythagorean Theorem to find any of the unknown information about the triangle. It is possible to use some right triangle trigonometry in combination with the Pythagorean Theorem to find the unknown information, but it's simpler to use the Law of Cosines—to find the unknown side and then to find the unknown angles.