Answer
(a) The error when computing the hypotenuse is $\pm 1.21~cm$
(b) The percentage error is $3.0\%$
Work Step by Step
Let $a$ be the side of length $20~cm$
Let $h$ be the hypotenuse.
Note that an error of $\pm 1^{\circ}$ is an error of $\pm \frac{\pi}{180}~rad$
We can find the error when computing the hypotenuse:
$\frac{a}{h} = sin~\theta$
$h = \frac{a}{sin~\theta}$
$dh = -\frac{a}{sin^2~\theta}~\cdot cos~\theta~d\theta$
$dh = -\frac{a~cos~\theta}{sin^2~\theta}~d\theta$
$dh = \pm~(\frac{20~cos~\frac{\pi}{6}}{sin^2~\frac{\pi}{6}}~\cdot \frac{\pi}{180})$
$dh = \pm~[\frac{20~\frac{\sqrt{3}}{2}}{(\frac{1}{2})^2}~\cdot \frac{\pi}{180}]$
$dh = \pm~\frac{40~\sqrt{3}~\pi}{180}$
$dh = \pm~1.21~cm$
The error when computing the hypotenuse is $\pm 1.21~cm$
(b) We can find the relative error:
$\frac{dh}{h} = \frac{-\frac{a~cos~\theta}{sin^2~\theta}~d\theta}{\frac{a}{sin~\theta}}$
$\frac{dh}{h} = -\frac{cos~\theta~d\theta}{sin~\theta}$
$\frac{dh}{h} = -cot~\theta~d\theta$
$\frac{dh}{h} = \pm (\frac{\pi}{180}~cot~\frac{\pi}{6})$
$\frac{dh}{h} = \pm 0.030$
The percentage error is $3.0\%$
