\documentclass{report} \input{preamble} \input{macros} \input{letterfonts} \title{\Huge{Calculus III}\\Homework \# 1} \author{\huge{Krishna Ayyalasomayajula}} \date{} \begin{document} \maketitle \newpage% or \cleardoublepage % \pdfbookmark[]{}{<dest>} \pdfbookmark[section]{\contentsname}{toc} \tableofcontents \pagebreak \chapter{Lab 2 - 13.1 Apps} \section{Work} \qs{}{ Let $\vec{T_1}$ represent the tension of the leftmost cable, while $\vec{T_2}$ encodes the tension force experienced by the rightmost cable. Our coordinate system will originate at the intersection of the two cables. \begin{align*} \vec{T_1} \coloneqq \langle \|\vec{T_1}\| ; 135\degree \rangle \\ \vec{T_2} \coloneqq \langle \|\vec{T_2}|\| ; 15\degree \rangle \\ 500=\|\vec{T_1}\|\sin{135\degree} + \|\vec{T_2}\|\sin{15} \\ 0 = \|\vec{T_1}\|\cos{135\degree} + \|\vec{T_2}\|\cos{15} \\ \implies 500=\|\vec{T_1}\|\tfrac{\sqrt{2}}{2} + \|\vec{T_2}\|\cdot0.258819045103 \\ \implies0 = \|\vec{T_1}\|\tfrac{-\sqrt{2}}{2} + \|\vec{T_2}\|\cdot0.965925826289 \\ \text{Solving the system numerically yields: }\\ \|\vec{T_1}\| \Rightarrow 557.677\, \mathrm{lb}\\ \|\vec{T_2}\| \Rightarrow 408.248\,\mathrm{lb} \\ \text{This corresponds to answer choice A} \end{align*} } \qs{}{ } \end{document}