Question

Find the measures of the three angles, in radians, of the triangle with the given vertices: d(1,1,1), e(1,−5,2), and f(−2,2,7).

Answer 1

Consider triangle DEF with vertices D(1,1,1), E(1,-5,2) and F(-2,2,7).

1. Find

$\overrightarrow{DE}=(1-1,-5-1,2-1)=(0,-6,1),\\ \\\overrightarrow{DF}=(-2-1,2-1,7-1)=(-3,1,6).$

Then

$\cos \angle D=\dfrac{0\cdot (-3)+(-6)\cdot 1+1\cdot 6}{\sqrt{0^2+(-6)^2+1^2}\cdot \sqrt{(-3)^2+1^2+6^2}}=\dfrac{0}{\sqrt{37} \cdot \sqrt{46} }=0.$

2. Find

$\overrightarrow{ED}=(1-1,1-(-5),1-2)=(0,6,-1),\\ \\\overrightarrow{EF}=(-2-1,2-(-5),7-2)=(-3,7,5).$

Then

$\cos \angle E=\dfrac{0\cdot (-2)+6\cdot 7+(-1)\cdot 5}{\sqrt{0^2+6^2+(-1)^2}\cdot \sqrt{(-3)^2+7^2+5^2}}=\dfrac{37}{\sqrt{37} \cdot \sqrt{83} }=\sqrt{\dfrac{37}{83}}.$

3. Find

$\overrightarrow{FE}=(1-(-2),-5-2,2-7)=(3,-7,-5),\\ \\\overrightarrow{FD}=(1-(-2),1-2,1-7)=(3,-1,-6).$

Then

$\cos \angle F=\dfrac{3\cdot 3+(-7)\cdot (-1)+(-5)\cdot (-6)}{\sqrt{3^2+(-7)^2+(-5)^2}\cdot \sqrt{3^2+(-1)^2+(-6)^2}}=\dfrac{46}{\sqrt{83} \cdot \sqrt{46} }=\sqrt{\dfrac{46}{83}}.$

4.

$\angle E=\arccos0=\dfrac{\pi}{2},\\ \\ \angle D=\arccos\letf(\sqrt{\dfrac{37}{83}}\right)\approx 0.27\pi,\\ \\ \angle F=\arccos\letf(\sqrt{\dfrac{46}{83}}\right)\approx 0.23\pi.$

Answer 2

The measures of the three angles are 90°, 48°, and 42°.

Triangle

Triangle is a polygon that has three sides and three angles. The sum of the angle of the triangle is 180 degrees.

Given

The triangle with the given vertices: d(1,1,1), e(1,−5,2), and f(−2,2,7).

To find

The measures of the three angles, in radians.

How to get the measures of the three angles, in radians?

We know the distance formula

$\rm Distance = \sqrt{(x_{2} -x_{1} )^{2} + (y_{2} -y_{1} )^{2} +(z_{2} -z_{1} )^{2} }$

Distance between DE will be

$\rm DE= \sqrt{(1-1 )^{2} + (-5-1 )^{2} +(2-1)^{2} } \\\\\rm DE= 6.083$

Distance between EF will be

$\rm EF= \sqrt{(-2-1 )^{2} + (2+5 )^{2} +(7-2)^{2} } \\\\\rm EF= 9.110$

Distance between DF will be

$\rm DF= \sqrt{(-2-1 )^{2} + (2-1 )^{2} +(7-1)^{2} } \\\\\rm DF= 6.782$

Then according to the cosine rule.

$\rm cos \theta = \dfrac{a^{2} +b^{2} -c^{2} }{2ab}$

Angle D will be

$\begin{aligned} \rm cos\ D &= \dfrac{6.782^{2} +6.083^{2} -9.110^{2} }{2*6.782*6.083}\\D &= 90 \\\end{aligned}$

Angle E will be

$\begin{aligned} \rm cos\ E &= \dfrac{9.110^{2} +6.083^{2} -6.782^{2} }{2*9.110*6.083}\\E &= 48\\\end{aligned}$

Then the sum of the angle of the triangle is 180 degrees.

 ∠D + ∠E + ∠F = 360°

90° + 48° + ∠F = 360°

                   ∠F = 42°

Thus the measures of the three angles are 90°, 48°, and 42°.

More about the triangle link is given below.

https://brainly.com/question/25813512