Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.3 - The Dot Product - Exercises 12.3 - Page 712: 13


$\angle A=63.435^{\circ}$ $\angle B=53.130^{\circ}$ $\angle C=63.435^{\circ}$

Work Step by Step

Angle at $A:$ Let ${\bf u}=\vec{AB}=\langle 2+1,1-0\rangle=\langle 3,1\rangle$, ${\bf v}=\vec{AC}=\langle 1+1,-2-0\rangle=\langle 2,-2\rangle$, $\displaystyle \angle A=\cos^{-1}(\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|})=\cos^{-1}\frac{(3)(2)+(1)(-2)}{\sqrt{3^{2}+1^{2}}\cdot\sqrt{2^{2}+(-2)^{2}}}$ $=\displaystyle \cos^{-1}\frac{4}{\sqrt{10}\cdot\sqrt{8}}=\cos^{-1}\frac{1}{\sqrt{5}}$ $=$ calculator gives = $63.435^{\circ}$ Angle at $B:$ Let ${\bf u}=\vec{BC}=\langle 1-2,-2-1\rangle=\langle-1,-3\rangle$, ${\bf v}=\vec{BA}=-\vec{AB}=\langle-3,-1\rangle$ $\displaystyle \angle B=\cos^{-1}(\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|})=\cos^{-1}(\frac{(-1)(-3)+(-3)(-1)}{\sqrt{10}\cdot\sqrt{10}})$ $=\displaystyle \cos^{-1}\frac{6}{10}=\cos^{-1}(0.6)$ $= $calculator gives = $53.130^{\circ}$ $\angle C=180^{\circ}-(\angle A+\angle B) =180^{\circ}-116.565^{\circ}$= $63.435^{\circ}$
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