Answer:
77.47°
75.96°
26.57°
Step-by-step explanation:
Given vertices of the triangle:
Find the vectors from A to B, B to C and A to C:
[tex]\begin{aligned}AB = B - A &=(x_B-x_A,y_B-y_A) \\&=(-3-(-2), 8-4)\\& = (-1, 4) \end{aligned}[/tex]
[tex]\begin{aligned}BC=C-B &=(x_C-x_B,y_C-y_B)\\ &=(6-(-3),8-8)\\&=(9,0)\end{aligned}[/tex]
[tex]\begin{aligned}AC = C - A &=(x_C-x_A,y_C-x_A)\\&= (6-(-2), 8-4) \\&= (8, 4)\end{aligned}[/tex]
Use Pythagoras Theorem to calculate the magnitudes of the vectors:
[tex]|AB| = \sqrt{(-1)^2+4^2}=\sqrt{17}[/tex]
[tex]|BC|=\sqrt{9^2+0^2}=9[/tex]
[tex]|AC| = \sqrt{8^2+4^2}=4\sqrt{5}[/tex]
[tex]\boxed{\begin{minipage}{6 cm}\underline{Dot Product of two vectors}\\\\$a \cdot b=|a||b| \cos \theta$\\\\where:\\ \phantom{ww}$\bullet$ $|a|$ is the magnitude of vector a. \\ \phantom{ww}$\bullet$ $|b|$ is the magnitude of vector b. \\ \phantom{ww}$\bullet$ $\theta$ is the angle between $a$ and $b$. \\ \end{minipage}}[/tex]
Rearrange the dot product formula to make θ the subject:
[tex]\implies \theta=\cos^{-1}\left(\dfrac{a \cdot b}{|a||b|}\right)[/tex]
Use the rearranged dot product formula to find the angles between two pairs of vectors.
[tex]\boxed{\begin{minipage}{4 cm}\underline{Dot Product}\\\\$\textbf{u} \cdot \textbf{v}=u_1v_1+u_2v_2$\\\\where:\\ \phantom{ww}$\bullet$ $\textbf{u}=\left\langle u_1,u_2 \right\rangle$ \\\phantom{ww}$\bullet$ $\textbf{v}= \left\langle v_1,v_2 \right\rangle$ \\ \end{minipage}}[/tex]
Angle A
[tex]\implies A=\cos^{-1}\left(\dfrac{AB \cdot AC}{|AB||AC|}\right)[/tex]
[tex]\implies A=\cos^{-1}\left(\dfrac{-1 \cdot 8+4 \cdot4}{\sqrt{17} \cdot 4 \sqrt{5}}\right)[/tex]
[tex]\implies A=\cos^{-1}\left(\dfrac{8}{4 \sqrt{85}}\right)[/tex]
[tex]\implies A=77.47^{\circ}\; \sf (2 \; d.p.)[/tex]
Angle C
[tex]\implies C=\cos^{-1}\left(\dfrac{BC \cdot AC}{|BC||AC|}\right)[/tex]
[tex]\implies C=\cos^{-1}\left(\dfrac{9 \cdot 8+0 \cdot4}{9 \cdot 4 \sqrt{5}}\right)[/tex]
[tex]\implies C=\cos^{-1}\left(\dfrac{72}{36 \sqrt{5}}\right)[/tex]
[tex]\implies C=26.57^{\circ}\; \sf (2 \; d.p.)[/tex]
Interior angles of a triangle sum to 180°.
[tex]\implies B=180^{\circ}-A-C[/tex]
[tex]\implies B=180^{\circ}-77.47^{\circ}-26.57^{\circ}[/tex]
[tex]\implies B=75.96^{\circ}[/tex]
Therefore, the interior angles of the triangle with the given vertices are:
