went back to the flashy styles

labs-set-1/template.tex

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+\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[<level>]{<title>}{<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}