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 $Q(3,5, - 1)$:
\[

$$\begin{align*} PQ&=\sqrt{(3+3)^2+(5 - 3)^2+(-1 - 4)^2}\\ &=\sqrt{6^2+2^2+(-5)^2}\\ &=\sqrt{36 + 4+25}\\ &=\sqrt{65}\approx 8.062 \end{align*}$$

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

$$\begin{align*} QR&=\sqrt{(1 - 3)^2+(0 - 5)^2+(2 + 1)^2}\\ &=\sqrt{(-2)^2+(-5)^2+3^2}\\ &=\sqrt{4 + 25+9}\\ &=\sqrt{38}\approx6.164 \end{align*}$$

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

$$\begin{align*} RP&=\sqrt{(-3 - 1)^2+(3 - 0)^2+(4 - 2)^2}\\ &=\sqrt{(-4)^2+3^2+2^2}\\ &=\sqrt{16 + 9+4}\\ &=\sqrt{29}\approx5.385 \end{align*}$$

\]

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 $PR$):
\[

$$\begin{align*} \cos\angle PQR&=\frac{PQ^{2}+QR^{2}-RP^{2}}{2\cdot PQ\cdot QR}\\ &=\frac{65 + 38-29}{2\sqrt{65}\sqrt{38}}\\ &=\frac{74}{2\sqrt{65\times38}}\\ &\approx0.697 \end{align*}$$

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

$$\begin{align*} \cos\angle QRP&=\frac{QR^{2}+RP^{2}-PQ^{2}}{2\cdot QR\cdot RP}\\ &=\frac{38 + 29-65}{2\sqrt{38}\sqrt{29}}\\ &=\frac{2}{2\sqrt{38\times29}}\\ &\approx0.038 \end{align*}$$

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

$$\begin{align*} \cos\angle RPQ&=\frac{RP^{2}+PQ^{2}-QR^{2}}{2\cdot RP\cdot PQ}\\ &=\frac{29+65 - 38}{2\sqrt{29}\sqrt{65}}\\ &=\frac{56}{2\sqrt{29\times65}}\\ &\approx0.697 \end{align*}$$

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

Answer:

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