Answer
$51.91^{\circ}$
Work Step by Step
Let the points be A(10,8), B(0,10) and C(3,2).
We calculate the vectors as follows: $\vec{AB}=⟨0-10,10-8⟩=⟨-10,2⟩$ $\vec{AC}=⟨3-10,2-8⟩=⟨-7,-6⟩$
Next, compute the lengths: $||\vec{AB}||=\sqrt {(-10)^{2}+2^{2}}=\sqrt {104}=2\sqrt {26}$ $||\vec{AC}||=\sqrt {(-7)^{2}+(-6)^{2}}=\sqrt {85}$
Now, calculate the dot product: $\vec{AB}\cdot\vec{AC}=⟨-10,2⟩\cdot⟨-7,-6⟩$ $=-10\times(-7)+2\times-6=58$
But $\vec{AB}\cdot\vec{AC}=||\vec{AB}||\,||\vec{AC}||\cos\theta$, where $\theta$ is the angle between $\overline{AB}$ and $\overline{AC}$.
$\implies 58=2\sqrt {26}\times\sqrt {85}\times\cos\theta$
Or,
$\theta=\cos^{-1}(\frac{58}{2\sqrt {26}\times\sqrt {85}})=51.91^{\circ}$
