Answer
$39.25 \mathrm{ft}^{2}$
Work Step by Step
We are given that the area $(A)$ of triangle where $a$ and b are the lengths of the two sides and $\theta$ is the included angle:
\[
\frac{1}{2} a b \sin \theta=A
\]
So:
\[
(a \sin \theta d b+b \sin \theta d a+a b \cos \theta d \theta)\frac{1}{2}=d A
\]
Both sides are divided by area:
\[
\frac{1}{2}\left(\frac{d \theta}{\tan \theta}+\frac{d b}{b}+\frac{d a}{a}\right)=\frac{d A}{A}
\]
Given the error in $\theta$ is very small, we can write $\tan \theta$ as $\theta$
\[
\begin{array}{l}
\frac{d A}{A}\left|=\frac{1}{2}\left(\left|\frac{d \theta}{\theta}+\frac{d b}{b}+\frac{d a}{a}\right|\right)\right. \\
\frac{d A}{A}\left|\leq \frac{1}{2}\left(\left|\frac{d \theta}{\theta}\right|+\left|\frac{d b}{b}\right|+\left|\frac{d a}{a}\right|\right)\right.
\end{array}
\]
We are given the maximum errors in a, b and $\theta$ are
\[
\left|\frac{d \theta}{\theta}\right|=2^{\circ}=\frac{\pi}{90}
\]
\[
\begin{array}{l}
\frac{1}{4} =\left|\frac{d b}{b}\right|\\
\frac{1}{2} =\left|\frac{d a}{a}\right|\\
\frac{d A}{A} | \leq \frac{1}{2}\left(\frac{\pi}{90}+\frac{1}{4}+\frac{1}{2}\right) \\
\frac{d A}{A} | \leq 0.3925
\end{array}
\]
$39.25 \mathrm{ft}^{2}$
