The coordinates of the vertices of the triangle A (2, 3, 4), B (3, 4, 4), C (4, 4, 6) are given.

The coordinates of the vertices of the triangle A (2, 3, 4), B (3, 4, 4), C (4, 4, 6) are given. Then the inner angle of the triangle at the vertex A is ..

Let’s find the coordinates of the vectors:

AB = (1; 1; 0);

BC = (1; 0; 2);

AC = (2; 1; 2).

Then their modules are equal to the sides of the triangle:

AB = √1 ^ 2 + 1 ^ 2 + 0 = √2;

BC = √1 ^ 2 + 2 ^ 2 = √5;

AC = √2 ^ 2 + 1 ^ 2 + 2 ^ 2 = √9 = 3.

By the cosine theorem, we get:

(√2) ^ 2 + 3 ^ 3 – 2√2 * 3 cos (A) = (√5) ^ 2;

-6√2cos (A) = 25 – 4 = 16;

cos (A) = – 16 / 6√2.

Answer: the desired angle is arccos (- 16 / 6√2).