Respuesta :
Answer:
The three angles are 43.11°, 103.95° and 32.94°.
Step-by-step explanation:
Let us name the corresponding angles to A, B and C to be ∠a, ∠b and ∠c respectively as shown in the figure below.
Using the vector representation of lines, we get,
Vector AB = < 1-5, 0+4, -1-0 > = < -4, 4, -1 >
Vector AC = < 1-5, 4+4, 5-0 > = < -4, 8, 5 >
Also, we have the modulus value given by,
|AB| = [tex]\sqrt{(-4)^{2}+4^{2}+(-1)^{2}}[/tex] = [tex]\sqrt{33}[/tex] = 5.75
|AC| = [tex]\sqrt{(-4)^{2}+8^{2}+5^{2}}[/tex] = [tex]\sqrt{105}[/tex] = 10.25
Now, using the dot product, we have,
AB · AC = |AB| |AC| cos(a)
[tex]\cos a=\frac{< -4,4,-1> < -4,8,5> }{5.75 \times 10.25}[/tex]
i.e. [tex]\cos a=\frac{16+32-5}{5.75 \times 10.25}[/tex]
i.e. [tex]\cos a=\frac{43}{58.9}[/tex]
i.e. [tex]\cos a=0.73[/tex]
i.e. [tex]a=\arccos 0.73[/tex]
i.e. [tex]a=43.11[/tex]
Hence, ∠a = 43.11°
Again, we see that,
Vector BA = < 5-1, -4-0, 0+1 > = < 4, -4, 1 >
Vector BC = < 1-1,4-0, 5+1 > = < 0, 4, 6 >
Also, we have the modulus value given by,
|Ba| = [tex]\sqrt{(4)^{2}+(-4)^{2}+1^{2}}[/tex] = [tex]\sqrt{33}[/tex] = 5.75
|BC| = [tex]\sqrt{0^{2}+4^{2}+6^{2}}[/tex] = [tex]\sqrt{52}[/tex] = 7.21
Now, using the dot product, we have,
BA · BC = |BA| |BC| cos(b)
[tex]\cos b=\frac{< 4,-4,1> < 0,4,6> }{5.75 \times 7.21}[/tex]
i.e. [tex]\cos b=\frac{0-16+6}{5.75 \times 7.21}[/tex]
i.e. [tex]\cos b=\frac{-10}{41.46}[/tex]
i.e. [tex]\cos b=-0.241[/tex]
i.e. [tex]b=\arccos -0.241[/tex]
i.e. [tex]b=103.95[/tex]
Hence, ∠b = 103.95°
Since, the sum of all the angles in a triangle is 180°.
Thus, ∠a + ∠b + ∠c = 180°
i.e. 43.11° + 103.95° + ∠c = 180°
i.e. ∠c = 180° - 147.06°
i.e. ∠c = 32.94°
Hence, the three angles are 43.11°, 103.95° and 32.94°.
Correct to the nearest degree, the three angles of the triangle are [tex]43.11^\circ[/tex], [tex]103.95^\circ[/tex], and [tex]32.94^\circ[/tex] and this can be determined by using the properties of the triangle and dot product formula.
Given :
The vertices of triangle: a(1, 0, −1), b(5, −4, 0), c(1, 4, 5)
Let the angles are [tex]\alpha[/tex], [tex]\beta[/tex], and [tex]\theta[/tex]. Now, the magnitude of AB will be:
[tex]\rm |AC| = \sqrt{(1-5)^2+(0+4)^2+(-1-0)^2}= \sqrt{33} = 5.75[/tex]
Now, the magnitude of AC will be:
[tex]\rm |AC| = \sqrt{(1-5)^2+(4+4)^2+(5-0)^2}= \sqrt{105} = 10.25[/tex]
Using dot product for finding the angles:
[tex]\rm cos \alpha =\dfrac{AB.AC}{|AB||AC|}[/tex]
[tex]\rm cos \alpha =\dfrac{16+32-5}{5.75\times 10.25}[/tex]
[tex]\rm cos \alpha =\dfrac{43}{58.9}[/tex]
[tex]\alpha = 43.11^\circ[/tex]
Now, the magnitude of BA will be:
[tex]\rm |BA| = \sqrt{(5-1)^2+(-4-0)^2+(0+1)^2}= \sqrt{33} = 5.75[/tex]
Now, the magnitude of BC will be:
[tex]\rm |BC| = \sqrt{(1-1)^2+(4-0)^2+(5+1)^2}= \sqrt{52} = 7.21[/tex]
Using dot product for finding the angles:
[tex]\rm cos \beta =\dfrac{BA.BC}{|BA||BC|}[/tex]
[tex]\rm cos \beta =\dfrac{0-16+6}{5.75\times 7.21}[/tex]
[tex]\rm cos \beta =\dfrac{-10}{41.46}[/tex]
[tex]\rm cos \beta =-0.241[/tex]
[tex]\beta = 103.95^\circ[/tex]
Now, the sum of interior angles of the triangle is [tex]180^\circ[/tex], then:
[tex]103.95^\circ+43.11^\circ+\theta = 180^\circ[/tex]
[tex]\theta = 32.94^\circ[/tex]
For more information, refer to the link given below:
https://brainly.com/question/14455586
