 Riley Barton

2023-03-29

Find, correct to the nearest degree, the three angles of the triangle with the given vertices
A(1, 0, -1), B(3, -2, 0), C(1, 3, 3) progutannx7f

To find the angles of the triangle with the given vertices, we can use the dot product formula to calculate the angle between two vectors. The dot product of two vectors is given by the formula:
$𝐮·𝐯=|𝐮|·|𝐯|·\mathrm{cos}\left(\theta \right)$
where $𝐮$ and $𝐯$ are the vectors, $|𝐮|$ and $|𝐯|$ are their magnitudes, and $\theta$ is the angle between them.
Let's calculate the angles of the triangle:
1. Angle A: To find angle A, we need to calculate the vectors $\stackrel{\to }{AB}$ and $\stackrel{\to }{AC}$, and then find the angle between them. Using the given vertices, we have:
$\stackrel{\to }{AB}=⟨3-1,-2-0,0-\left(-1\right)⟩=⟨2,-2,1⟩$
$\stackrel{\to }{AC}=⟨1-1,3-0,3-\left(-1\right)⟩=⟨0,3,4⟩$
Now, we can calculate the dot product:
$\stackrel{\to }{AB}·\stackrel{\to }{AC}=\left(2\right)\left(0\right)+\left(-2\right)\left(3\right)+\left(1\right)\left(4\right)=-6+4=-2$
Next, we calculate the magnitudes:
$|\stackrel{\to }{AB}|=\sqrt{{2}^{2}+\left(-2{\right)}^{2}+{1}^{2}}=\sqrt{9}=3$
$|\stackrel{\to }{AC}|=\sqrt{{0}^{2}+{3}^{2}+{4}^{2}}=\sqrt{25}=5$
Now, we can find the angle A:
$\mathrm{cos}\left(A\right)=\frac{\stackrel{\to }{AB}·\stackrel{\to }{AC}}{|\stackrel{\to }{AB}|·|\stackrel{\to }{AC}|}=\frac{-2}{3·5}=-\frac{2}{15}$
Taking the inverse cosine, we get:
$A={\mathrm{cos}}^{-1}\left(-\frac{2}{15}\right)\approx {97.7}^{\circ }$
2. Angle B: To find angle B, we follow the same steps as above, but using the vectors $\stackrel{\to }{BA}$ and $\stackrel{\to }{BC}$. We have:
$\stackrel{\to }{BA}=-\stackrel{\to }{AB}=⟨-2,2,-1⟩$
$\stackrel{\to }{BC}=⟨1-3,3-\left(-2\right),3-0⟩=⟨-2,5,3⟩$
Calculating the dot product:
$\stackrel{\to }{BA}·\stackrel{\to }{BC}=\left(-2\right)\left(-2\right)+\left(2\right)\left(5\right)+\left(-1\right)\left(3\right)=4+10-3=11$
Calculating the magnitudes:
$|\stackrel{\to }{BA}|=|\stackrel{\to }{AB}|=3$
$|\stackrel{\to }{BC}|=\sqrt{\left(-2{\right)}^{2}+{5}^{2}+{3}^{2}}=\sqrt{38}$
Finding the angle B:
$\mathrm{cos}\left(B\right)=\frac{\stackrel{\to }{BA}·\stackrel{\to }{BC}}{
|\stackrel{\to }{BA}|·|\stackrel{\to }{BC}|}=\frac{11}{3·\sqrt{38}}$

Taking the inverse cosine:
$B={\mathrm{cos}}^{-1}\left(\frac{11}{3·\sqrt{38}}\right)\approx {53.5}^{\circ }$
3. Angle C: To find angle C, we can use the fact that the sum of the angles in a triangle is 180 degrees. Therefore, we can calculate angle C by subtracting angles A and B from 180 degrees:
$C=180-A-B=180-97.7-53.5\approx {28.8}^{\circ }$
Therefore, the angles of the triangle are approximately:
Angle A: 97.7 degrees
Angle B: 53.5 degrees
Angle C: 28.8 degrees

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