diff --git a/labs-set-1/template.pdf b/labs-set-1/template.pdf
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diff --git a/labs-set-1/template.tex b/labs-set-1/template.tex
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--- a/labs-set-1/template.tex
+++ b/labs-set-1/template.tex
@@ -39,6 +39,76 @@
 }
 
 \qs{}{
+  Let $\vec{T}$ represent the tension in the cable, and $\vec{C}$ represent the compression force experienced by the boom in neutralizing the horizontal component of $\vec{T}$.
 
+  \begin{align*}
+    \vec{T}\coloneqq \langle \|\vec{T}\| ; 142\degree \rangle \\
+    \|\vec{T}\|\sin{142\degree} = 450 \, \mathrm{lb} \\
+    \|\vec{T}\| = \frac{450 \, \mathrm{lb}}{\sin{142\degree}} \\
+    \|\vec{T}\| \Rightarrow 730.921 \, \mathrm{lb} \\
+    \|\vec{C}\| = |(\|\vec{T}\|\cos{142\degree})|\Rightarrow 575.973 \, \mathrm{lb} \\ 
+    \text{ This corresponds to answer choice C}
+  \end{align*}
 }
+
+\qs{}{
+  Let the $x'$-axis of the new coordinate system be oriented along the unit vector $\hat{u}=\hat{T_1}$:
+  \begin{align*}
+    \vec{T_1} \coloneqq \langle 3600,0 \rangle \\ 
+    \vec{T_2} \coloneqq \langle 1800;45\degree \rangle \\
+  \|\vec{T_1}+\vec{T_2}\| = \langle 3600+1800\cos{45\degree},1800\sin{45\degree} \rangle = \langle 3600+1800\tfrac{\sqrt{2}}{2},1800\tfrac{\sqrt{2}}{2} \rangle \\
+  \|\vec{T_1}+\vec{T_2}\| \Rightarrow 5036.278 \, \mathrm{lb} \\
+  \text{This corresponds with answer choice D}
+  \end{align*}
+}
+
+
+\qs{}{
+  Let the force vector be $\vec{F}$. 
+
+  \begin{align*}
+    \vec{F} \coloneqq 8\cdot\cos{30\degree}\hat i + 8\cdot\sin{30\degree}\hat j = 8\cdot\tfrac{\sqrt{3}}{2}\hat i + 8\cdot\tfrac{1}{2}\hat j  = 4\cdot\sqrt{3} \hat i + 4 \hat j\\
+    \text{This corresponds with answer choice A}
+  \end{align*}
+}
+
+\qs{}{
+  Let the velocity vector be $\vec{v_0}$. 
+
+  \begin{align*}
+    \vec{v_0} \coloneqq 5\cdot\cos{56\degree}\hat i + 5\cdot\sin{56\degree}\hat j  = 2.795 \hat i + 4.145 \hat j\\
+    \text{This corresponds with answer choice A}
+  \end{align*}
+}
+
+\qs{}{
+  Let the velocity vector be $\vec{v_0}$. 
+
+  \begin{align*}
+    \vec{v_0} \coloneqq \langle 817 ; 140\degree \rangle = \langle 817\cos{140\degree},817\sin{140\degree} \rangle \Rightarrow \langle -625.858,525.157 \rangle \\
+    \text{This corresponds with answer choice A}
+  \end{align*}
+}
+
+\qs{}{
+  Let the wind's parameters be encoded in the vector $\vec{W}$. 
+
+  \begin{align*}
+    \vec{W}\coloneqq\langle 5,14 \rangle \\
+  \implies \theta = \arctan{\tfrac{14}{5}} \Rightarrow 70.346 \degree \\
+    \text{This corresponds with answer choice C}
+  \end{align*}
+}
+\pagebreak
+\qs{}{
+  Let the boat's target velocity vector be $\vec{v}$. 
+
+  \begin{align*}
+    \vec{v}\coloneqq \langle 0,31 \rangle - \langle -6,0 \rangle \\
+    \vec{v} = \langle 6,31 \rangle \\
+    \implies \theta = \arctan{\tfrac{31}{6}} \Rightarrow 79.04593\degree \\ 
+    \text{This corresponds with answer choice C, when expressed as a bearing East of North}
+  \end{align*}
+}
+
 \end{document}