Question

Use vectors to find the interior angles (in degrees) of the triangle with the given vertices. (Round your answers to two decimal places.)

Answer 1

1. Find the vector between each pair of vertices:

To make it more easy give to each vertex a letter:

A(1,3)

B(3,5)

C(2,6)

$\begin{gathered} \vec{AB}=(3,5)-(1,3) \\ \vec{AB}=(3-1,5-3) \\ \vec{AB}=(2,2) \end{gathered}$$\begin{gathered} \vec{BC}=(2,6)-(3,5) \\ \vec{BC}=(2-3,6-5) \\ \vec{BC}=(-1,1) \end{gathered}$$\begin{gathered} \vec{CA}=(1,3)-(2,6) \\ \vec{CA}=(1-2,3-6) \\ \vec{CA}=(-1,-3) \end{gathered}$

2. Find the magnitude of each vector:

$\begin{gathered} \lvert{AB}\rvert=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8} \\ \lvert{BC}\rvert=\sqrt{(-1)\placeholder{⬚}^2+1^2}=\sqrt{1+1}=\sqrt{2} \\ \lvert{CA}\rvert=\sqrt{(-1)\placeholder{⬚}^2+(-3)\placeholder{⬚}^2}=\sqrt{1+9}=\sqrt{10} \end{gathered}$

3. Fid the dot product beween each pair of vectors:

$\begin{gathered} \vec{AB}\cdot\vec{BC}=2*-1+2*1=-2+2=0 \\ \vec{AB}\cdot\vec{CA}=2*-1+2*-3=-2-6=-8 \\ \vec{CA}\cdot\vec{BC}=-1*-1+-3*1=-1-3=-4 \end{gathered}$

4. Use the next formula to find the angle betweeen two vectors:

$\theta=\cos^{-1}(\frac{a\cdot b}{\sqrt{a}\cdot\sqrt{b}})$

Angle between AB and BC:

$\begin{gathered} \theta=\cos^{-1}(\frac{0}{\sqrt{8}\cdot\sqrt{2}}) \\ \\ \theta=\cos^{-1}(0) \\ \\ \theta=90º \end{gathered}$

Angle between AB and CA: