Question

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

Answer 1

Answer:

angle CAB = 113.8 degree

angle ABC = 35.6 degree

angle BCA = 30.6 degree

Step-by-step explanation:

Given data:

A(1, 0, −1),

B(5, −3, 0),

C(1, 5, 2)

calculate the length of side by using the distance formula

so

AB = (5,-3,0) - (1,0,-1) = (4,-3,1)

AC= (1,5,2) - (1,0,-1) = (0,5,3)

|AB|

|AC| =$\sqrt {(0 + 5^2+3^2)} = \sqrt{34}$

From following formula, calculate the angle between the two side i.e Ab and AC

AB.AC = |AB|*|AC| cos ∠CAB

(4,-3,1).(0,5,3)

4*0 -3*5 +1*3

-12 =

cos ∠CAB = - 0.404

angle CAB = 113.8 degree

BA =B- A =  (1,0,-1) - (5,-3,0) = (-4,3,-1)

BC = (1,5,2)-(5,-3,0) = (-4,8,2)

|BA| = \sqrt{(26)}

|BC| $= \sqrt {(4^2 + 8^2 + 2^2)} = \sqrt{(84)}$

BA.BC = |BA|*|BC|* cosABC

(-4,3,-1).(-4,8,2) =$\sqrt{(26)} * \sqrt{(84)} *cosABC$

16+24-2

cos ∠ABC = 0.813

angle ABC = 35.6 degree

we know sum of three angle in a traingle is 180 degree hence

sum of all three angle = 180

angle BCA + 35.6 + 113.8 = 180

angle BCA = 30.6 degree