There is a problem: A right triangle has a base $b = 100\pm 1 \text{ ft}$ and adjacent angle $\theta = 30^\circ\pm 0.5^\circ$. Calculate the height $h$ and its uncertainty.
Thus, for the height, I did this: $h = b \tan(\theta) = (100) tan (30^\circ)= 57.74\ldots \text{ ft}$
for the uncertainty. I set up this: $\sigma_h = \sqrt{\left(\frac{\partial h}{\partial \theta}\right)^2 (\sigma_\theta)^2 + \left(\frac{\partial h}{\partial b}\right)^2 (\sigma_b)^2}$
I calculated $$\frac{\partial h}{\partial \theta}=b \sec^2(\theta)$$ and $$\frac{\partial h}{\partial b}= \tan^2(\theta)$$
Plugging back into the equation:
$$\sigma_h =\sqrt{(100 \sec^2(30^\circ))^2 (0.5)^2 + (\tan^2(30^\circ))^2 (1)^2}$$
which gave me $66.66 \ldots $
My teacher said I'm suppose to get around $10\text{ ft}$ for the uncertainty though.
I can't figure out what I'm doing wrong in this particular problem. Please help!
You started with $$h=b \tan(\theta)$$ So $$\frac{dh}{db}=\tan(\theta)$$ $$\frac{dh}{d\theta}=b \sec ^2(\theta )$$ $$\sigma_h^2=\Big(\frac{dh}{db}\Big)^2\sigma_b^2+\Big(\frac{dh}{d\theta }\Big)^2\sigma_{\theta}^2=b^2 \sec ^4(\theta )\sigma_{\theta}^2+\tan ^2(\theta )\sigma_b^2$$
Now, with your numbers : $\theta=\frac{\pi}{6}$,$b=100$, $\sigma_b=1$, $\sigma_{\theta}=\frac{\pi}{360}$, after some basic simplifications $$\sigma_h^2=\frac{1}{\sqrt{3}}+\frac{\pi ^2}{972}$$
I think that your problem is related to the derivative of any trigonometric function when the angle is in degrees. If I may suggest, use radians everywhere.