Hey, I'm not sure if I got this answer correct, I've been having a hard time with using differentials to find the various types of errors, I was hoping someone could look over this and point me in the right direction if I'm way off.

The question isSo I know that the relative error is found through the quotient relation $\displaystyle \frac{\triangle A}{A}\approx\frac{dA}{A}=\frac{(A')(d\theta)}{A}$, so to solve this, I plug the differentiated function $\displaystyle A'$ and the given function $\displaystyle A$ in, as well as $\displaystyle d\theta=\frac{0.25}{30}=0.008333$ right? *I got my $\displaystyle d\theta$ value from the ratio between the number of erroneous minutes per whole degree by the total number of degrees.The area of a right triangle with hypotenuse $\displaystyle H$ is given by the formula

$\displaystyle A=\frac{1}{4}H^2sin(2\theta)$

where $\displaystyle \theta$ is one of the acute angles. Use differentials to estimate the relative errors of the area$\displaystyle A$ if $\displaystyle H=4 cm$ exactly and $\displaystyle \theta$ is measured to be $\displaystyle 30^o$ with an error of measurement of $\displaystyle \pm15'$

Is this the correct path to take? My work to solve the problem is below.

$\displaystyle \frac{\triangle A}{A}\approx\frac{dA}{A}=\frac{(A')(d\theta)}{A}$

$\displaystyle \frac{(A')(d\theta)}{A}=\frac{[(\frac{1}{2}H)(\sin(2\theta))+(\frac{1}{4}H^2)(\co s(2\theta)2)](0.008333)}{\frac{1}{4}H^2\sin(2\theta)}$

$\displaystyle \frac{(A')(d\theta)}{A}=\frac{(2\sin(60)+4\cos(60) )(0.008333)}{4\sin(60)}$

$\displaystyle \frac{(A')(d\theta)}{A}=\frac{1.732+4}{3.464}$

$\displaystyle \frac{(A')(d\theta)}{A}=0.01373=\boxed{1.373\%}$

So the relative error is $\displaystyle 1.373\%$?

Thanks for taking the time to look this over!

-Kasper