Question

what are the angles of the triangle with vertices p(-3, 3, 4), q(3, 5, -1) and r(1, 0, 2)? enter your answer as three angles, separated by commas. enter all angles in degrees, round to one decimal place. enter \deg\ for the degree symbol.

Explanation:

Step1: Calculate the side - lengths using the distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For side $PQ$ with $P(-3,3,4)$ and $Q(3,5, - 1)$:

$$
LATEXBLOCK0
$$

For side $QR$ with $Q(3,5,-1)$ and $R(1,0,2)$:

$$
LATEXBLOCK1
$$

For side $RP$ with $R(1,0,2)$ and $P(-3,3,4)$:

$$
LATEXBLOCK2
$$

Step2: Use the cosine - law to find the angles

The cosine - law states that $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}$, where $a,b,c$ are the side - lengths of the triangle and $A$ is the angle opposite to side $a$.
For angle $\angle PQR$ (opposite to side $PR$):

$$
LATEXBLOCK3
$$

$\angle PQR=\cos^{-1}(0.697)\approx45.8^{\circ}$
For angle $\angle QRP$ (opposite to side $QP$):

$$
LATEXBLOCK4
$$

$\angle QRP=\cos^{-1}(0.038)\approx87.8^{\circ}$
For angle $\angle RPQ$ (opposite to side $RQ$):

$$
LATEXBLOCK5
$$

$\angle RPQ=\cos^{-1}(0.697)\approx45.8^{\circ}$

Answer:

$45.8$ deg,$87.8$ deg,$45.8$ deg