# Math Help - Percentage error in Area of triangle

1. ## Percentage error in Area of triangle

Hello, I'm stuck on the following question:

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?

2. ## Re: Percentage error in Area of triangle

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

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

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

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

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

- Hollywood