Percentage error: area of triangle

Any help would be fantastic.

Thanks

I am having problems with the following question:

"The area of a triangle ABC is calculated using the formula:

S=1/2bcsinA

and it is known that b, c and A are measured correctly to within 1%. If the angle A is measured as 45 degrees, prove that the percentage error in the calculated value of S is not more than about 2.8%."

When I use the equation:

$\Delta$S= $\displaystyle \frac{\partial S}{\partial b}$$\Delta$b + $\displaystyle \frac{\partial S}{\partial c}$ $\Delta$c + $\displaystyle \frac{\partial S}{\partial A}$$\Delta$A

and subsequently divide by S, I get:

$\Delta$S/S= $\Delta$b/b + $\Delta$c/c + (cosA/sinA) $\Delta$A

and at this point I am totally stuck. What happens to the $\Delta$A here? And for the cosA/sinA do I use angle A as 45 degrees?

Any help would be fantastic.

Thanks

The only thing I was doing wrong was I was using degrees instead of radians. I got the correct answer. Thanks!