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=√(2^2+2^2)=√(4+4)=√(8) \\ \lvert{BC}\rvert=\sqrt{(-1)\placeholder{⬚}^2+1^2}=√(1+1)=√(2) \\ \lvert{CA}\rvert=\sqrt{(-1)\placeholder{⬚}^2+(-3)\placeholder{⬚}^2}=√(1+9)=√(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)((a\cdot b)/(√(a)\cdot√(b)))`

Angle between AB and BC:

`\begin{gathered} \theta=\cos^(-1)((0)/(√(8)\cdot√(2))) \\ \\ \theta=\cos^(-1)(0) \\ \\ \theta=90º \end{gathered}`

Angle between AB and CA: