Question

use vectors to find the interior angles of the triangle given the following sets of vertices. round your answer, in degrees, to two decimal places. (5, -8), (8, 9), (-9, -7)

Explanation:

Step1: Label the points and find vectors

Let \(A=(5,-8)\), \(B=(8,9)\), \(C = (-9,-7)\)
Vector \(\overrightarrow{AB}=\langle8 - 5,9-(-8)
angle=\langle3,17
angle\)
Vector \(\overrightarrow{AC}=\langle-9 - 5,-7-(-8)
angle=\langle-14,1
angle\)
Vector \(\overrightarrow{BA}=\langle5 - 8,-8 - 9
angle=\langle-3,-17
angle\)
Vector \(\overrightarrow{BC}=\langle-9 - 8,-7 - 9
angle=\langle-17,-16
angle\)
Vector \(\overrightarrow{CA}=\langle5+9,-8 + 7
angle=\langle14,-1
angle\)
Vector \(\overrightarrow{CB}=\langle8 + 9,9+7
angle=\langle17,16
angle\)

Step2: Use the dot - product formula \(\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\vert\vec{u}\vert\vert\vec{v}\vert}\)

For angle \(A\):
\(\vec{u}=\overrightarrow{AB}=\langle3,17
angle\), \(\vec{v}=\overrightarrow{AC}=\langle-14,1
angle\)
\(\vec{u}\cdot\vec{v}=(3)\times(-14)+(17)\times(1)=-42 + 17=-25\)
\(\vert\vec{u}\vert=\sqrt{3^{2}+17^{2}}=\sqrt{9 + 289}=\sqrt{298}\)
\(\vert\vec{v}\vert=\sqrt{(-14)^{2}+1^{2}}=\sqrt{196 + 1}=\sqrt{197}\)
\(\cos A=\frac{-25}{\sqrt{298}\sqrt{197}}\approx\frac{-25}{17.26\times14.04}\approx - 0.103\)
\(A=\cos^{-1}(-0.103)\approx95.91^{\circ}\)

For angle \(B\):
\(\vec{u}=\overrightarrow{BA}=\langle-3,-17
angle\), \(\vec{v}=\overrightarrow{BC}=\langle-17,-16
angle\)
\(\vec{u}\cdot\vec{v}=(-3)\times(-17)+(-17)\times(-16)=51+272 = 323\)
\(\vert\vec{u}\vert=\sqrt{(-3)^{2}+(-17)^{2}}=\sqrt{9 + 289}=\sqrt{298}\)
\(\vert\vec{v}\vert=\sqrt{(-17)^{2}+(-16)^{2}}=\sqrt{289+256}=\sqrt{545}\)
\(\cos B=\frac{323}{\sqrt{298}\sqrt{545}}\approx\frac{323}{17.26\times23.34}\approx0.803\)
\(B=\cos^{-1}(0.803)\approx36.57^{\circ}\)

For angle \(C\):
Since the sum of angles in a triangle is \(180^{\circ}\)
\(C=180-(A + B)\)
\(C=180-(95.91+36.57)=47.52^{\circ}\)

Answer:

The interior angles of the triangle are approximately \(A = 95.91^{\circ}\), \(B = 36.57^{\circ}\), \(C = 47.52^{\circ}\)