So in the answer of this question i want to understand two point first one is this how he change \begin{align} (a^2+b^2-2ab\cos(\angle AOB))=a^2b^2\sin^2(\angle AOB)\\ \end{align} the second point and it's so important to me is how he solve (8) for cos(∠AOB) using the quadratic formula to get this equation
\begin{align} \cos(\angle AOB)=\frac{p^2-\sqrt{(p^2-c^2a^2)(p^2-b^2c^2)}}{abc^2}\ \end{align}
i would really someone to help me here
thanks
He's using two different formulas for the area of a triangle and equating them. One formula is $$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$$ or $$|\triangle AOB|= \frac{1}{2}|\overline{AB}||\overline{OF}|\tag {1}$$ The other formula is the side-angle-side formula which gives $$|\triangle AOB|= \frac{1}{2}ab\sin(\angle AOB)\tag{2}$$ He then inserts formulas for $|\overline{AB}|$ and $|\overline{OF}|$ in equation $(1)$, squares both sides and replaces $\sin^2(\angle AOB)$ with ($1-\cos^2(\angle AOB)$) to get $$(a^2+b^2-2ab\cos(\angle AOB))(p/c)^2 = a^2b^2(1-\cos^2(\angle AOB))$$ The above is a quadratic equation with the variable $\cos(\angle AOB)$, which he then solves.