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Percentage error in Area of triangle

Hello, I'm stuck on the following question:

Attachment 26204

I only get as far as taking natural logs and then differentiation it, but have no idea on how to carry on:

ln(S) = ln(b)+ln(c)+ln(sinA)

Assuming that b and c are equal to 1, I get:

1/S dS = 1/b db + 1/c dc + cot A (sub in [A=pi/4])

That's what I gather from the text book but then I hit dead end.

Any idea on what I'm doing wrong or whether I'm completely wrong from the off?

Re: Percentage error in Area of triangle

You are given that $\displaystyle \frac{1}{b}\ db$ and $\displaystyle \frac{1}{c}\ dc$ are both 5%. For the third term, you correctly differentiated

$\displaystyle d(\log(\sin{A}))=(\cot{A})dA$

but you are given $\displaystyle \frac{dA}{A}$, so you need:

$\displaystyle d(\log(\sin{A}))=(\cot{A})A\ \frac{dA}{A}$.

So $\displaystyle \frac{dA}{A}$ is 5%, A is $\displaystyle \frac{\pi}{4}$ and $\displaystyle \cot{A}$ is 1.

- Hollywood