Question

e vectors to find the interior angles of the triangle given the following sets of vertices. Round your answer, in degrees, to two decimal places.(6, 1). (8, -4), (-9, -6)

Answer 1

`\begin{gathered} Angle\text{ at \lparen-9, -6\rparen is }18.31^(\circ) \\ Angle\text{ at \lparen8, -4\rparen is }74.91^(\circ) \\ Angle\text{ at \lparen6, 1\rparen is }86.14^(\circ) \end{gathered}`

Step-by-step explanation:

Given:

`A(6,1),B(8,-4),C(-9,-6)`

To find:

The interior angles of the triangle using vectors

If we draw this triangle using the given vertices, we'll have;

To find the angle at C(-9, -6) we need to get the vector name for vectors CA and CB as seen below;

`\begin{gathered} CA<(6-(-9),1-(-6)>\rightarrow CA<15,7> \\ CB<8-(-9),-4-(-6)>\rightarrow CA<17,2> \end{gathered}`

Then we'll the below formula to solve for the angle at C;

`\begin{gathered} \cos\theta=\frac{CA\cdot CB}{Magnitude\text{ of CA and CB}} \\ \cos\theta=(255+14)/(√(15^2+7^2)*√(17^2+2^2)) \\ \cos\theta=(269)/(√(274)*√(293)) \\ \cos\theta=(27)/(283.34) \\ \cos\theta=0.9494 \\ \theta=\cos^(-1)(0.9494) \\ \theta=18.31^(\circ) \end{gathered}`

So the angle at (-9, -6) is 18.31 degrees

To find the angle at B(8, -4) we need to get the vector name for vectors BA and BC as seen below;

`\begin{gathered} BA<(6-8,1-(-4)>\rightarrow AB<-2,5> \\ BC<-9-8,-6-(-4)>\rightarrow CA<-17,-2> \end{gathered}`

Then we'll the below formula to solve for the angle at B;

`\begin{gathered} \cos\theta=\frac{BA\cdot BC}{Magnitude\text{ of BA and BC}} \\ \cos\theta=(34+(-10))/(√((-2)^2+5^2)*√((-17)^2+(-2)^2)) \\ \cos\theta=(24)/(√(29)*√(293)) \\ \cos\theta=(24)/(92.1792) \\ \cos\theta=0.2604 \\ \theta=\cos^(-1)(0.2604) \\ \theta=74.91^(\circ) \\ So\text{ the angle at \lparen8, -4\rparen is 74.91 degrees} \end{gathered}`