Chapter 6.2, Problem 30E

Solution

To Find:

The interior angles of the triangle with given vertices.

  ${41.93}^{°}$ , ${74.20}^{°}$ and ${63.87}^{°}$ approximately.

Given:

The points $\left(-4,1\right)$ , $\left(1,-6\right)$ , and, $\left(5,-1\right)$ .

Concepts Used:

For the points $A=\left(x_1,y_1\right)$ and $B=\left(x_2,y_2\right)$ , the vector $\overrightarrow{AB}$ represents the vector with initial point $A$ and terminal point $B$ . The vector $\overrightarrow{AB}$ is represented in component form as $〈x_2-x_1,y_2-y_1〉$ . The vector $\overrightarrow{BA}$ represents the reversed vector and is given by $\overrightarrow{BA}=-\overrightarrow{AB}$ .

If $u=〈a,b〉$ then $\left|u\right|=\sqrt{a^2+b^2}$ .

The dot product of the vectors $u=〈u_1,u_2〉$ and $v=〈v_1,v_2〉$ is defined as:

  $u\cdot v=u_1{v}_1+u_2{v}_2$

The smaller angle $\theta$ between two vectors $u$ and $v$ is given by ${\cos }^{-1}\left(\frac{u\cdot v}{\left|u\right|\left|v\right|}\right)$

Calculations:

Label the points as $A=\left(-4,1\right)$ , $B=\left(1,-6\right)$ , and, $C=\left(5,-1\right)$ .

Determine the vector $\overrightarrow{AB}$ .

  $\begin{array}{c}\overrightarrow{AB}=〈1-\left(-4\right),-6-1〉\\ =〈1+4,-7〉\\ =〈5,-7〉\end{array}$

Determine the vector $\overrightarrow{AC}$ .

  $\begin{array}{c}\overrightarrow{AC}=〈5-\left(-4\right),-1-1〉\\ =〈5+4,-2〉\\ =〈9,-2〉\end{array}$

The interior angle $A$ is the angle between the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ . Determine the angle between the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ .

  $\begin{array}{c}\angle A={\cos }^{-1}\left(\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{\left|\overrightarrow{AB}\right|\left|\overrightarrow{AC}\right|}\right)\\ ={\cos }^{-1}\left(\frac{〈5,-7〉\cdot 〈9,-2〉}{\left|〈5,-7〉\right|\left|〈9,-2〉\right|}\right)\\ ={\cos }^{-1}\left(\frac{\left(5\right)\left(9\right)+\left(-7\right)\left(-2\right)}{\sqrt{{\left(5\right)}^2+{\left(-7\right)}^2}\sqrt{{\left(9\right)}^2+{\left(-2\right)}^2}}\right)\\ ={\cos }^{-1}\left(\frac{45+14}{\sqrt{25+49}\sqrt{81+4}}\right)\\ ={\cos }^{-1}\left(\frac{59}{\sqrt{74}\sqrt{85}}\right)\\ \approx {41.93}^{°}\end{array}$

Determine the vector $\overrightarrow{BA}$ .

  $\begin{array}{c}\overrightarrow{BA}=-\overrightarrow{AB}\\ =-〈5,-7〉\\ =〈-5,7〉\end{array}$

Determine the vector $\overrightarrow{BC}$ .

  $\begin{array}{c}\overrightarrow{BC}=〈5-1,-1-\left(-6\right)〉\\ =〈4,-1+6〉\\ =〈4,5〉\end{array}$

The interior angle $B$ is the angle between the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$ . Determine the angle between the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$ .

  $\begin{array}{c}\angle B={\cos }^{-1}\left(\frac{\overrightarrow{BA}\cdot \overrightarrow{BC}}{\left|\overrightarrow{BA}\right|\left|\overrightarrow{BC}\right|}\right)\\ ={\cos }^{-1}\left(\frac{〈-5,7〉\cdot 〈4,5〉}{\left|〈-5,7〉\right|\left|〈4,5〉\right|}\right)\\ ={\cos }^{-1}\left(\frac{\left(-5\right)\left(4\right)+\left(7\right)\left(5\right)}{\sqrt{{\left(-5\right)}^2+{\left(7\right)}^2}\sqrt{{\left(4\right)}^2+{\left(5\right)}^2}}\right)\\ ={\cos }^{-1}\left(\frac{-20+35}{\sqrt{25+49}\sqrt{16+25}}\right)\\ ={\cos }^{-1}\left(\frac{15}{\sqrt{74}\sqrt{41}}\right)\\ \approx {74.20}^{°}\end{array}$

Determine the vector $\overrightarrow{CA}$ .

  $\begin{array}{c}\overrightarrow{CA}=-\overrightarrow{AC}\\ =-〈9,-2〉\\ =〈-9,2〉\end{array}$

Determine the vector $\overrightarrow{CB}$ .

  $\begin{array}{c}\overrightarrow{CB}=-\overrightarrow{BC}\\ =-〈4,5〉\\ =〈-4,-5〉\end{array}$

The interior angle $C$ is the angle between the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ . Determine the angle between the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ .

  $\begin{array}{c}\angle C={\cos }^{-1}\left(\frac{\overrightarrow{CA}\cdot \overrightarrow{CB}}{\left|\overrightarrow{CA}\right|\left|\overrightarrow{CB}\right|}\right)\\ ={\cos }^{-1}\left(\frac{〈-9,2〉\cdot 〈-4,-5〉}{\left|〈-9,2〉\right|\left|〈-4,-5〉\right|}\right)\\ ={\cos }^{-1}\left(\frac{\left(-9\right)\left(-4\right)+\left(2\right)\left(-5\right)}{\sqrt{{\left(-9\right)}^2+{\left(2\right)}^2}\sqrt{{\left(-4\right)}^2+{\left(-5\right)}^2}}\right)\\ ={\cos }^{-1}\left(\frac{36-10}{\sqrt{81+4}\sqrt{16+25}}\right)\\ ={\cos }^{-1}\left(\frac{26}{\sqrt{85}\sqrt{41}}\right)\\ \approx {63.87}^{°}\end{array}$

Conclusion:

The interior angle at the point $A=\left(-4,1\right)$ is $\approx {41.93}^{°}$ .

The interior angle at the point $B=\left(1,-6\right)$ is $\approx {74.20}^{°}$ .

The interior angle at the point $C=\left(5,-1\right)$ is $\approx {63.87}^{°}$ .