The area of a triangle ABC is given by S = 0.5abSin(C) calculate the max percentage error of S

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$S=0.5abSin(c)$ calculate the maximum percentage error of S when a,b and c have a percentage error within 1% and c = 45 degrees.

I do not understand where I went wrong the answer is supposed to 2.8%. (Using total differential).

$ \Delta s = \frac {\partial S}{\partial a} \Delta a + \frac {\partial S}{\partial b} \Delta b + \frac {\partial S}{\partial c} \Delta c $

where:

$\frac {\partial S}{\partial a}= \frac{1}{2} bSin(c)$,

$\frac {\partial S}{\partial b}= \frac{1}{2} aSin(c)$,

$\frac {\partial S}{\partial a}= \frac{1}{2} abCos(c)$

therefore:

$\Delta s = \frac{1}{2} bSin(c) + \frac{1}{2} aSin(c) + \frac{1}{2} abCos(c)$

hence:

$ \frac{ \Delta s}{s} = \frac {\frac{1}{2} bSin(c) + \frac{1}{2} aSin(c) + \frac{1}{2} abCos(c)}{ \frac{1}{2} absin(c) }$

$ \frac{\Delta s}{s} = \frac{\delta a}{a} + \frac{\delta b}{b} + cot(c) \Delta(c)$

now since: $\frac{\Delta(c)}{45} = 0.01$ that implies $\Delta(c) = 0.45$

Hence perc. error = 1 + 1 + 45 = 47% (which is incorrect)

Anyone know what I did wrong?

Thank you ;)

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You need to express $C$ in radians, not degrees. So $C=\pi/4$, and therefore $\Delta C=.01 (\frac\pi4)\approx .008$.