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 $A:{97.7}^{\circ }\phantom{\rule{1em}{0ex}}B:{53.5}^{\circ }\phantom{\rule{1em}{0ex}}C:{28.8}^{\circ }$

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