If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are sides of a triangle then what will the range of sum of cosines of angles of a triangle. I know that the range is $(1,\frac{3}{2})$. I can approach the answer using trigonometry but have no idea how to get this result using vectors.
someone please tell me what to do. Please give me a solution.
On
Here's a full solution:
Use the identity $\cos A+\cos B+\cos C =1+\frac{r}{R}$ where $r$ is the inradius and $R$ is the circumradius. The Euler inequality is $$\frac{R}{r}\ge 2\implies 1+\frac{r}{R}\le \frac{3}{2}$$ Clearly, $$\frac{r}{R}> 0$$ Which is where the lower bound comes from. You can show it's tight pretty easily.
The upper bound can be obtained as follows:
Let $O$ be the origin of vectors, $\vec{OA}, \vec{OB}, \vec{OC}$ three unit vectors such that $\angle (\vec{OA}, \vec{OB}) = \alpha, \angle(\vec{OB}, \vec{OC}) = \beta$. Then \begin{align*} (\vec{OA} + \vec{OB} + \vec{OC})^2 &\geq 0 \\ 3 + 2\cos \alpha + 2\cos \beta + 2\cos(\alpha + \beta) \geq 0 \end{align*} and hence $$\cos \alpha + \cos \beta + \cos(\alpha + \beta) \geq -\frac{3}{2}$$ Since this is true for all $\alpha, \beta$, letting $\alpha = 180^\circ - A, \beta = 180^\circ - B$, we get \begin{align*} -\cos A - \cos B + \cos(360^\circ - (A+B)) &\geq -\frac{3}{2} \\ -\cos A - \cos B + \cos(180^\circ + C) &\geq -\frac{3}{2} \\ -\cos A - \cos B -\cos C &\geq -\frac{3}{2} \\ \cos A + \cos B + \cos C &\leq \frac{3}{2} \end{align*}