Find, correct to the nearest degree, the three angles of the triangle with the given vertices. a(1, 0, −1), b(5, −4, 0), c(1, 4, 5)

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| = = = 5.75

|AC| = = = 10.25

Now, using the dot product, we have,

AB · AC = |AB| |AC| cos(a)

i.e.

i.e.

i.e.

i.e.

i.e.

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| = = = 5.75

|BC| = = = 7.21

Now, using the dot product, we have,

BA · BC = |BA| |BC| cos(b)

i.e.

i.e.

i.e.

i.e.

i.e.

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°.

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)