Pre-calculus Complex coordinate law of cosine help

Question

Let $\theta = \angle BAC$. Then we can write $\cos \theta = \dfrac{x}{\sqrt{2}}$ Find x.

Currently, that is all I have. I think I should use Law of cosine. Where should I progress?

Answer 1

Assuming that

$A=(1,1), B=(2,3), C=(4,0)$,

we know that $\vec{AB}=(2-1)\vec{x}+(3-1)\vec{y}=\vec{x}+2\vec{y}$, and $\vec{AC}=(4-1)\vec{x}+(0-1)\vec{y}=3\vec{x}-\vec{y}$.

Thus, $||\vec{AB}||=\sqrt{1^2+2^2}=\sqrt{5}$, and $||\vec{AC}||=\sqrt{3^2+(-1)^2}=\sqrt{10}$. Therefore, $||\vec{AB}||||\vec{AC}||=\sqrt{5}\sqrt{10}=5\sqrt{2}$.

And also: $\vec{AB}\cdot\vec{AC}=(1\cdot3)+(-1\cdot2)=3-2=1$.

Therefore, using a rule about dot products, $\cos\theta=\frac{\vec{AB} \cdot\vec{AC}}{||\vec{AB}||||\vec{AC}||}$

Plugging in, we get $\cos\theta=\frac{1}{5\sqrt{2}}$. Therefore, $x=\frac{1}{5}$

Answer 2

$$a = BC = \sqrt{2^2+3^2}=\sqrt{13}$$ $$b = AC = \sqrt{3^2+1^2}=\sqrt{10}$$ $$c = AB = \sqrt{2^2+1^2}=\sqrt{5}$$

so $$\cos \theta = {b^2+c^2-a^2\over 2bc} = {1\over 5\sqrt{2}}\implies x=1/5$$