# Find, correct to the nearest degree, the three angles of the triangle with the given vertices. A(1, 0, -1), B(3, -2, 0), C(1, 3, 3).

The main objective of this question is to find the three angles of a triangle given three vertices. The angles can be found using the dot product of the vectors representing the sides of the triangle.

A triangle is a polygon with three-sides that is also referred to as a trigon. Every triangle possesses $3$ sides and $3$ angles, which may or may not be the same. Triangles are classified as acute, equilateral, isosceles, obtuse, isosceles right, and right triangle.

A triangle is formed geometrically by the intersection of three line segments. In each triangle, every side has $2$ endpoints, and the endpoints of all three sides may intersect at three different points in a plane to form a triangle. The three intersecting points are referred to as triangle vertices. The angles inside a triangle are referred to as the interior angles and the sum of three angles of the triangle is always equal to $180^\circ$. Any triangle that is not a right triangle is defined as an oblique triangle.

Given vertices are:

$A(1, 0, -1), B(3, -2, 0), C(1, 3, 3)$

First, find the vectors representing the sides of the triangle.

$\overrightarrow{AB}=\langle 3-1,-2-0,0+1\rangle$ $=\langle 2,-2,1\rangle$

$\overrightarrow{AC}=\langle 1-1, 3-0,3+1\rangle$ $=\langle 0,3,4\rangle$

$\overrightarrow{BC}=\langle 1-3, 3+2,3-0\rangle$ $=\langle -2,5,3\rangle$

The magnitudes of the sides of triangle are:

$|\overrightarrow{AB}|=\sqrt{(2)^2+(-2)^2+(1)^2}$ $=3$

$|\overrightarrow{AC}|=\sqrt{(0)^2+(3)^2+(4)^2}$ $=5$

$|\overrightarrow{BC}|=\sqrt{(-2)^2+(5)^2+(3)^2}$ $=\sqrt{38}$

Let $\alpha$ be the angle between $\overrightarrow{AB}$ and $\overrightarrow{AC}$, then by using the dot product:

$\cos \alpha=\dfrac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|}$

$\cos \alpha=\dfrac{(2)(0)+(-2)(2)+(1)(4)}{(3)(5)}$

$\cos \alpha=\dfrac{0-4+4}{15}=$ $-\dfrac{2}{15}$

$\alpha=\cos^{-1}\left(-\dfrac{2}{15}\right)$

$\alpha=97.67^\circ$

Let $\beta$ be the angle between $\overrightarrow{AB}$ and $\overrightarrow{BC}$, then by using the dot product:

$\cos \beta=\dfrac{\overrightarrow{AB}\cdot\overrightarrow{BC}}{|\overrightarrow{AB}||\overrightarrow{BC}|}$

$\cos \beta=\dfrac{(2)(-2)+(-2)(5)+(1)(3)}{(3)(\sqrt{38})}$

$\cos \beta=\dfrac{-4-10+3}{3\sqrt{38}}=$ $-\dfrac{11}{3\sqrt{38}}$

$\beta=\cos^{-1}\left(-\dfrac{11}{3\sqrt{38}}\right)$

$\beta=126.5^\circ$

This is the angle outside the triangle because the direction $\overrightarrow{BC}$ is pointing relative to $\overrightarrow{AB}$, and so, we should find the supplementary angle that is:

$\beta=180^\circ-126.5^\circ$ $=53.5^\circ$

Let $\gamma$ be the angle between $\overrightarrow{AC}$ and $\overrightarrow{BC}$. Since the sum of the angles of a triangle is $180^\circ$, so:

$\alpha+\beta+\gamma=180^\circ$

$97.67^\circ+53.5^\circ+\gamma=180^\circ$

$151.17^\circ+\gamma=180^\circ$

$\gamma=180^\circ-151.17^\circ$

$\gamma=28.83^\circ$

## Example

Given the vertices $a(0,0),b(1,2),c(-1,4)$, solve for the three angles of a triangle.

### Solution

Given vertices are:

$a(0,0),b(1,2),c(-1,4)$

First, find the vectors representing the sides of the triangle.

$\overrightarrow{ab}=\langle 1-0,2-0\rangle$ $=\langle 1,2\rangle$

$\overrightarrow{ca}=\langle -1-0, 4-0\rangle$ $=\langle -1,4\rangle$

$\overrightarrow{bc}=\langle -1-1, 4-2\rangle$ $=\langle -2,2\rangle$

The magnitudes of the sides of triangle are:

$|\overrightarrow{ab}|=\sqrt{(1)^2+(2)^2}$ $=\sqrt{5}$

$|\overrightarrow{ca}|=\sqrt{(-1)^2+(4)^2}$ $=\sqrt{17}$

$|\overrightarrow{bc}|=\sqrt{(-2)^2+(2)^2}$ $=2\sqrt{2}$

Let $\alpha$ be the angle between $\overrightarrow{ab}$ and $\overrightarrow{ca}$, then by using the dot product:

$\cos \alpha=\dfrac{\overrightarrow{ab}\cdot\overrightarrow{ca}}{|\overrightarrow{ab}||\overrightarrow{ca}|}$

$\cos \alpha=\dfrac{(1)(-1)+(4)(2)}{(\sqrt{5})(\sqrt{17})}$

$\cos \alpha=\dfrac{-1-8}{\sqrt{85}}=$ $-\dfrac{9}{\sqrt{85}}$

$\alpha=\cos^{-1}\left(-\dfrac{9}{\sqrt{85}}\right)$

$\alpha=12.53^\circ$

Let $\beta$ be the angle between $\overrightarrow{ab}$ and $\overrightarrow{bc}$, then by using the dot product:

$\cos \beta=\dfrac{\overrightarrow{ab}\cdot\overrightarrow{bc}}{|\overrightarrow{ab}||\overrightarrow{bc}|}$

$\cos \beta=\dfrac{(1)(-2)+(2)(2)}{(\sqrt{5})(\sqrt{2})}$

$\cos \beta=\dfrac{-2+4}{\sqrt{10}}=$ $\dfrac{2}{\sqrt{10}}$

$\beta=\cos^{-1}\left(\dfrac{2}{\sqrt{10}}\right)$

$\beta=50.77^\circ$

Let $\gamma$ be the angle between $\overrightarrow{ca}$ and $\overrightarrow{bc}$. Since the sum of the angles of a triangle is $180^\circ$, so:

$\alpha+\beta+\gamma=180^\circ$

$12.53^\circ+50.77^\circ+\gamma=180^\circ$

$63.3^\circ+\gamma=180^\circ$

$\gamma=180^\circ-63.3^\circ$

$\gamma=116.7^\circ$

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