Suppose you have a triangle (not a right triangle) wher one angle is (theta). the length of the opposite to theta is 5. the other two sides have length 4 and 6.
so i have all the sides. picturing it like a normal triangle.. 5 is on the bottom, for theta is at the tip of the traingle with 4 and 6 on either side...now how can i find the angle theta? Theta would be the angle correct?
+26400 $$a=4\quad b=6\quad c=5 \quad \mbox{ angle } \theta\mbox{ ? }$$
$$\begin{array}{l}
c^2=a^2+b^2-2ab*\cos{(\theta) } \\\\
\cos{(\theta)} = \dfrac{a^2+b^2-c^2}{2ab}\\\\
\cos{(\theta)} = \dfrac{4^2+6^2-5^2}{2*4*6}\\\\
\cos{(\theta)} = \dfrac{16+36-25}{48}\\\\
\cos{(\theta)} = \dfrac{27}{48}\\\\
\cos{(\theta)} = 0.5625\\\\
\textcolor[rgb]{1,0,0}{\theta=55.7711336722\ensuremath{^\circ}}
\end{array}$$
+26400 $$a=4\quad b=6\quad c=5 \quad \mbox{ angle } \theta\mbox{ ? }$$
$$\begin{array}{l}
c^2=a^2+b^2-2ab*\cos{(\theta) } \\\\
\cos{(\theta)} = \dfrac{a^2+b^2-c^2}{2ab}\\\\
\cos{(\theta)} = \dfrac{4^2+6^2-5^2}{2*4*6}\\\\
\cos{(\theta)} = \dfrac{16+36-25}{48}\\\\
\cos{(\theta)} = \dfrac{27}{48}\\\\
\cos{(\theta)} = 0.5625\\\\
\textcolor[rgb]{1,0,0}{\theta=55.7711336722\ensuremath{^\circ}}
\end{array}$$
