+26393 How can you find the angles of a triangle if you have all the sides?
\(\begin{array}{|rcll|} \hline \cos{(B)} &=& \frac{1}{2c}\cdot \left(a+\frac{c^2-b^2}{a} \right) \\ \cos{(C)} &=& \frac{1}{2b}\cdot \left(a-\frac{c^2-b^2}{a} \right) \\ A &=& 180^{\circ}-(B+C) \\ \hline \end{array}\)
+502 You can use the Cosine rule which is
CosA= c2 + b2- a2/ 2bc
and the same with B and C
+23252 If the angles of the triangle are A, B, and C with the opposite sides a, b, and c,
then C = cos-1( (a2 + b2 - c2) / (2·a·b) ).
Similar formulas for the other angles.
+26393 How can you find the angles of a triangle if you have all the sides?
\(\begin{array}{|rcll|} \hline \cos{(B)} &=& \frac{1}{2c}\cdot \left(a+\frac{c^2-b^2}{a} \right) \\ \cos{(C)} &=& \frac{1}{2b}\cdot \left(a-\frac{c^2-b^2}{a} \right) \\ A &=& 180^{\circ}-(B+C) \\ \hline \end{array}\)
