Find, correct to the nearest degree, the three angles of the triangle with the given vertices.

Asked by maham237 @ 16/12/2020 in Mathematics viewed by 351 People

a(1, 0, −1), b(5, −4, 0), c(1, 4, 5)

Answered by maham237 @ 16/12/2020

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| = $\sqrt{(-4)^{2}+4^{2}+(-1)^{2}}$ = $\sqrt{33}$ = 5.75

|AC| = $\sqrt{(-4)^{2}+8^{2}+5^{2}}$ = $\sqrt{105}$ = 10.25

Now, using the dot product, we have,

AB · AC = |AB| |AC| cos(a)

$\cos a=\frac{< -4,4,-1> < -4,8,5> }{5.75 \times 10.25}$

i.e. $\cos a=\frac{16+32-5}{5.75 \times 10.25}$

i.e. $\cos a=\frac{43}{58.9}$

i.e. $\cos a=0.73$

i.e. $a=\arccos 0.73$

i.e. $a=43.11$

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| = $\sqrt{(4)^{2}+(-4)^{2}+1^{2}}$ = $\sqrt{33}$ = 5.75

|BC| = $\sqrt{0^{2}+4^{2}+6^{2}}$ = $\sqrt{52}$ = 7.21

Now, using the dot product, we have,

BA · BC = |BA| |BC| cos(b)

$\cos b=\frac{< 4,-4,1> < 0,4,6> }{5.75 \times 7.21}$

i.e. $\cos b=\frac{0-16+6}{5.75 \times 7.21}$

i.e. $\cos b=\frac{-10}{41.46}$

i.e. $\cos b=-0.241$

i.e. $b=\arccos -0.241$

i.e. $b=103.95$

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°.

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