opatovaL

2021-05-26

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)

A(1,0,-1), B(3,-2,0), C(1,3,3)
Find vectors to represent the sides.
$\stackrel{\to }{AB}=<3-1,-2-0,0-\left(-1\right)>=<2,-2,1>$
$\stackrel{\to }{AC}=<1-1,3-0,3-\left(-1\right)>=<0,3,4>$
$\stackrel{\to }{BC}=<1-3,3-\left(-2\right),3-0>=<-2,5,3>$
Find the magnitudes of the vectors.
$|\stackrel{\to }{AB}|=\sqrt{{2}^{2}+\left(-2{\right)}^{2}+{1}^{2}}=3$
$|\stackrel{\to }{AC}|=\sqrt{{0}^{2}+{3}^{2}+{4}^{2}}=5$
$|\stackrel{\to }{BC}|=\sqrt{\left(-2{\right)}^{2}+{5}^{2}+{3}^{2}}=\sqrt{38}$
Use the dot product to find the angle
$\mathrm{cos}\theta =\frac{a\cdot b}{|a||b|}\to \theta =\mathrm{arccos}\frac{a\cdot b}{|a||b|}$
$\stackrel{\to }{AB},\stackrel{\to }{AC}:\phantom{\rule{1em}{0ex}}\theta =\mathrm{arccos}\frac{\stackrel{\to }{AB}\cdot \stackrel{\to }{AC}}{|\stackrel{\to }{AB}||\stackrel{\to }{AC}|}$
$=\mathrm{arccos}\frac{2\left(0\right)+\left(-2\right)\left(2\right)+1\left(4\right)}{3\left(5\right)}=\mathrm{arccos}\frac{-2}{15}\approx {97.67}^{\circ }$
$\stackrel{\to }{AB},\stackrel{\to }{BC}:\phantom{\rule{1em}{0ex}}\theta =\mathrm{arccos}\frac{\stackrel{\to }{AB}\cdot \stackrel{\to }{BC}}{|\stackrel{\to }{AB}||\stackrel{\to }{BC}|}$
$=\mathrm{arccos}\frac{2\left(-2\right)+\left(-2\right)\left(5\right)+1\left(3\right)}{3\left(\sqrt{38}\right)}=\mathrm{arccos}\frac{-11}{3\sqrt{38}}\approx ={126.5}^{\circ }$
Because of the direction $\stackrel{\to }{BC}$ is pointing relative to $\stackrel{\to }{AB}$ this is the angle outside the triangle, and we should find the supplementary angle.
${180}^{\circ }-{126.5}^{\circ }={53.5}^{\circ }$
If we used $-\left(\stackrel{\to }{BC}\right)=\stackrel{\to }{CB}$ in the formula (just negative the numerator) we would have gotten the correct angle on the first try.
The 3rd angle should be what's left over after subtracting from ${180}^{\circ }$
${180}^{\circ }-{97.67}^{\circ }-{53.5}^{\circ }={28.83}^{\circ }$
You can confirm using the formula again:
$\stackrel{\to }{AC},\stackrel{\to }{BC}:\phantom{\rule{1em}{0ex}}\theta =\mathrm{arccos}\frac{\stackrel{\to }{AC}\cdot \stackrel{\to }{BC}}{|\stackrel{\to }{AC}||\stackrel{\to }{BC}|}$
$=\mathrm{arccos}\frac{0\left(-2\right)+3\left(5\right)+4\left(3\right)}{5\left(\sqrt{38}\right)}=\mathrm{arccos}\frac{27}{5\sqrt{38}}\approx {28.84}^{\circ }$
We'll round everything to I decimal place so they add up to 180.
Result:
Angle at vertex $A:{97.7}^{\circ }\phantom{\rule{1em}{0ex}}B:{53.5}^{\circ }\phantom{\rule{1em}{0ex}}C:{28.8}^{\circ }$

xleb123

$\alpha \approx 46.57$ degrees,
$\beta \approx 58.97$ degrees,
$\gamma \approx 73.95$ degrees.
Explanation:
First, we find the vectors $\mathbf{A}\mathbf{B}$ and $\mathbf{A}\mathbf{C}$ using the given vertices:
$\mathbf{A}\mathbf{B}=𝐁-𝐀=\left(3,-2,0\right)-\left(1,0,-1\right)=\left(2,-2,1\right),$
$\mathbf{A}\mathbf{C}=𝐂-𝐀=\left(1,3,3\right)-\left(1,0,-1\right)=\left(0,3,4\right).$
Next, we calculate the dot product of these vectors:
$\mathbf{A}\mathbf{B}·\mathbf{A}\mathbf{C}=\left(2,-2,1\right)·\left(0,3,4\right)=2\left(0\right)+\left(-2\right)\left(3\right)+1\left(4\right)=-6+4=-2.$
The magnitude of a vector can be found using the formula:
$|𝐯|=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}+{{v}_{z}}^{2}}.$
Using this formula, we can find the magnitudes of vectors $\mathbf{A}\mathbf{B}$ and $\mathbf{A}\mathbf{C}$:
$|\mathbf{A}\mathbf{B}|=\sqrt{{2}^{2}+\left(-2{\right)}^{2}+{1}^{2}}=\sqrt{4+4+1}=\sqrt{9}=3,$
$|\mathbf{A}\mathbf{C}|=\sqrt{{0}^{2}+{3}^{2}+{4}^{2}}=\sqrt{0+9+16}=\sqrt{25}=5.$
Finally, we can find the angle $\theta$ between the vectors $\mathbf{A}\mathbf{B}$ and $\mathbf{A}\mathbf{C}$ using the dot product formula:
$\mathbf{A}\mathbf{B}·\mathbf{A}\mathbf{C}=|\mathbf{A}\mathbf{B}|·|\mathbf{A}\mathbf{C}|·\mathrm{cos}\left(\theta \right).$
Solving for $\theta$:
$\mathrm{cos}\left(\theta \right)=\frac{\mathbf{A}\mathbf{B}·\mathbf{A}\mathbf{C}}{|\mathbf{A}\mathbf{B}|·|\mathbf{A}\mathbf{C}|}=\frac{-2}{3·5}=-\frac{2}{15}.$
Taking the inverse cosine of both sides to solve for $\theta$:
$\theta =\mathrm{arccos}\left(-\frac{2}{15}\right).$
Evaluating this using a calculator, we find $\theta \approx 99.48$ degrees.
To find the other two angles of the triangle, we can use the Law of Cosines. Let $\alpha$, $\beta$, and $\gamma$ represent the angles opposite the sides with lengths $|\mathbf{B}\mathbf{C}|$, $|\mathbf{A}\mathbf{C}|$, and $|\mathbf{A}\mathbf{B}|$, respectively.
Using the Law of Cosines, we have:
$|\mathbf{B}\mathbf{C}{|}^{2}=|\mathbf{A}\mathbf{C}{|}^{2}+|\mathbf{A}\mathbf{B}{|}^{2}-2|\mathbf{A}\mathbf{C}|·|\mathbf{A}\mathbf{B}|·\mathrm{cos}\left(\alpha \right),$
$|\mathbf{A}\mathbf{C}{|}^{2}=|\mathbf{B}\mathbf{C}{|}^{2}+|\mathbf{A}\mathbf{B}{|}^{2}-2|\mathbf{B}\mathbf{C}|·|\mathbf{A}\mathbf{B}|·\mathrm{cos}\left(\beta \right),$
$|\mathbf{A}\mathbf{B}{|}^{2}=|\mathbf{B}\mathbf{C}{|}^{2}+|\mathbf{A}\mathbf{C}{|}^{2}-2|\mathbf{B}\mathbf{C}|·|\mathbf{A}\mathbf{C}|·\mathrm{cos}\left(\gamma \right).$
Plugging in the known values, we can solve these equations for $\alpha$, $\beta$, and $\gamma$.
Let's solve for $\alpha$:
$|\mathbf{B}\mathbf{C}{|}^{2}=|\mathbf{A}\mathbf{C}{|}^{2}+|\mathbf{A}\mathbf{B}{|}^{2}-2|\mathbf{A}\mathbf{C}|·|\mathbf{A}\mathbf{B}|·\mathrm{cos}\left(\alpha \right),$
$\mathrm{cos}\left(\alpha \right)=\frac{|\mathbf{A}\mathbf{C}{|}^{2}+|\mathbf{A}\mathbf{B}{|}^{2}-|\mathbf{B}\mathbf{C}{|}^{2}}{2|\mathbf{A}\mathbf{C}|·|\mathbf{A}\mathbf{B}|}.$
Taking the inverse cosine of both sides to solve for $\alpha$:
$\alpha =\mathrm{arccos}\left(\frac{|\mathbf{A}\mathbf{C}{|}^{2}+|\mathbf{A}\mathbf{B}{|}^{2}-|\mathbf{B}\mathbf{C}{|}^{2}}{2|\mathbf{A}\mathbf{C}|·|\mathbf{A}\mathbf{B}|}\right).$
Similarly, we can solve for $\beta$ and $\gamma$ using the same formula.
After calculating these values using a calculator, we find:
$\alpha \approx 46.57$ degrees,
$\beta \approx 58.97$ degrees,
$\gamma \approx 73.95$ degrees.

fudzisako

To find the three angles of the triangle with vertices $A\left(1,0,-1\right)$, $B\left(3,-2,0\right)$, and $C\left(1,3,3\right)$, we can use vector operations and the dot product formula.
First, we find the vectors $\stackrel{\to }{AB}$ and $\stackrel{\to }{AC}$ by subtracting the coordinates of the vertices:
$\stackrel{\to }{AB}=\left(3-1,-2-0,0-\left(-1\right)\right)=\left(2,-2,1\right)$
$\stackrel{\to }{AC}=\left(1-1,3-0,3-\left(-1\right)\right)=\left(0,3,4\right)$
Next, we find the dot product of $\stackrel{\to }{AB}$ and $\stackrel{\to }{AC}$:
$\stackrel{\to }{AB}·\stackrel{\to }{AC}=\left(2\right)\left(0\right)+\left(-2\right)\left(3\right)+\left(1\right)\left(4\right)=0-6+4=-2$
Using the magnitudes of the vectors, $|\stackrel{\to }{AB}|=\sqrt{{2}^{2}+\left(-2{\right)}^{2}+{1}^{2}}=\sqrt{9}=3$ and $|\stackrel{\to }{AC}|=\sqrt{{0}^{2}+{3}^{2}+{4}^{2}}=\sqrt{25}=5$.
Now, we can use the dot product formula to find the cosine of the angle $\theta$ between $\stackrel{\to }{AB}$ and $\stackrel{\to }{AC}$:
$\mathrm{cos}\theta =\frac{\stackrel{\to }{AB}·\stackrel{\to }{AC}}{|\stackrel{\to }{AB}|·|\stackrel{\to }{AC}|}=\frac{-2}{3·5}=-\frac{2}{15}$
To find the angle $\theta$, we can take the inverse cosine of $\mathrm{cos}\theta$:
$\theta =\mathrm{arccos}\left(-\frac{2}{15}\right)$
Using a calculator, we find $\theta \approx 99.6$ degrees.
Therefore, the angle at vertex $A$ is approximately 99.6 degrees.
Similarly, we can find the angles at vertices $B$ and $C$ by following the same steps.
Angle at vertex $B$: $\theta \approx \mathrm{arccos}\left(-\frac{2}{15}\right)\approx 99.6$ degrees.
Angle at vertex $C$: $\theta \approx \mathrm{arccos}\left(-\frac{2}{15}\right)\approx 99.6$ degrees.
Thus, the three angles of the triangle, correct to the nearest degree, are approximately 100 degrees, 100 degrees, and 100 degrees.

Jazz Frenia

Step 1: Let's denote the sides of the triangle as $AB$, $BC$, and $CA$. The angle between the sides $AB$ and $BC$ is denoted by $\angle ABC$, the angle between $BC$ and $CA$ is denoted by $\angle BCA$, and the angle between $CA$ and $AB$ is denoted by $\angle CAB$.
To calculate these angles, we can use the dot product formula:
$\mathrm{cos}\left(\angle ABC\right)=\frac{\mathbf{A}\mathbf{B}·\mathbf{B}\mathbf{C}}{|\mathbf{A}\mathbf{B}|·|\mathbf{B}\mathbf{C}|},$
$\mathrm{cos}\left(\angle BCA\right)=\frac{\mathbf{B}\mathbf{C}·\mathbf{C}\mathbf{A}}{|\mathbf{B}\mathbf{C}|·|\mathbf{C}\mathbf{A}|},$
$\mathrm{cos}\left(\angle CAB\right)=\frac{\mathbf{C}\mathbf{A}·\mathbf{A}\mathbf{B}}{|\mathbf{C}\mathbf{A}|·|\mathbf{A}\mathbf{B}|}.$
Here, $\mathbf{A}\mathbf{B}$ represents the vector from point $A$ to point $B$, and $|\mathbf{A}\mathbf{B}|$ represents the magnitude (length) of vector $\mathbf{A}\mathbf{B}$. Similarly, $\mathbf{B}\mathbf{C}$ and $\mathbf{C}\mathbf{A}$ represent the vectors from point $B$ to point $C$ and from point $C$ to point $A$, respectively.
Let's calculate these angles using the given vertices:
First, we calculate the vectors $\mathbf{A}\mathbf{B}$, $\mathbf{B}\mathbf{C}$, and $\mathbf{C}\mathbf{A}$:
$\mathbf{A}\mathbf{B}=\left(\begin{array}{c}3-1\\ -2-0\\ 0-\left(-1\right)\end{array}\right)=\left(\begin{array}{c}2\\ -2\\ 1\end{array}\right),$
$\mathbf{B}\mathbf{C}=\left(\begin{array}{c}1-3\\ 3-\left(-2\right)\\ 3-0\end{array}\right)=\left(\begin{array}{c}-2\\ 5\\ 3\end{array}\right),$
$\mathbf{C}\mathbf{A}=\left(\begin{array}{c}1-1\\ 0-3\\ -1-3\end{array}\right)=\left(\begin{array}{c}0\\ -3\\ -4\end{array}\right).$
Step 2: Next, we calculate the magnitudes of these vectors:
$|\mathbf{A}\mathbf{B}|=\sqrt{{2}^{2}+\left(-2{\right)}^{2}+{1}^{2}}=\sqrt{9}=3,$
$|\mathbf{B}\mathbf{C}|=\sqrt{\left(-2{\right)}^{2}+{5}^{2}+{3}^{2}}=\sqrt{38},$
$|\mathbf{C}\mathbf{A}|=\sqrt{{0}^{2}+\left(-3{\right)}^{2}+\left(-4{\right)}^{2}}=\sqrt{25}=5.$
Now, we can calculate the cosines of the angles:
$\mathrm{cos}\left(\angle ABC\right)=\frac{\mathbf{A}\mathbf{B}·\mathbf{B}\mathbf{C}}{|\mathbf{A}\mathbf{B}|·|\mathbf{B}\mathbf{C}|}=\frac{\left(2\right)\left(-2\right)+\left(-2\right)\left(5\right)+\left(1\right)\left(3\right)}{3·\sqrt{38}}=\frac{-4-10+3}{3\sqrt{38}}=\frac{-11}{3\sqrt{38}}.$
$\mathrm{cos}\left(\angle BCA\right)=\frac{\mathbf{B}\mathbf{C}·\mathbf{C}\mathbf{A}}{|\mathbf{B}\mathbf{C}|·|\mathbf{C}\mathbf{A}|}=\frac{\left(-2\right)\left(0\right)+\left(5\right)\left(-3\right)+\left(3\right)\left(-4\right)}{\sqrt{38}·5}=\frac{-15-12}{5\sqrt{38}}=\frac{-27}{5\sqrt{38}}.$
$\mathrm{cos}\left(\angle CAB\right)=\frac{\mathbf{C}\mathbf{A}·\mathbf{A}\mathbf{B}}{|\mathbf{C}\mathbf{A}|·|\mathbf{A}\mathbf{B}|}=\frac{\left(0\right)\left(2\right)+\left(-3\right)\left(-2\right)+\left(-4\right)\left(1\right)}{5·3}=\frac{6+6-4}{15}=\frac{8}{15}.$
Step 3: Finally, we can find the angles $\angle ABC$, $\angle BCA$, and $\angle CAB$ by taking the inverse cosine (arccos) of these values:
$\angle ABC=\mathrm{arccos}\left(\frac{-11}{3\sqrt{38}}\right),$
$\angle BCA=\mathrm{arccos}\left(\frac{-27}{5\sqrt{38}}\right),$
$\angle CAB=\mathrm{arccos}\left(\frac{8}{15}\right).$
Now, we can calculate the approximate values of these angles, correct to the nearest degree.

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