The central angle of a cirular sector is $80°$. It is desired to reduce it by $1°$. By how much should the radius of the sector be increased so that the area will remain unchanged if the original length of the radius is $20cm$?

Let $A$ be the area of the sector if function of the angle $\theta rad$ and the radius $R$. Then, we have that $A(\theta, R) = \frac{\theta R^2}{2} $. I want to aproximate my $\triangle A$, that is, the variation of the area, by

$$ dA = \frac{\partial A}{\partial \theta} d\theta + \frac{\partial A}{\partial R}dR$$

We have that $dA = 0$ and we want to find $dR$

I've found that:

• $80° = \frac{4 \pi}{9} rad$;

• $R = \frac{45}{\pi} cm$

•$1° = \frac{\pi}{180} rad$

Using this, I couldn't get the answer given ($1/8 cm$). Maybe I am thinking wrong.

Can you help me?

Thanks in advance!


1 Answer 1


Guess it's just a little error in your calculation. Start with keeping everything in cm and degrees; they are just units. $$ 0 = \frac{20^2}{2} \times (-1) + 20 \times 80 \times dR $$ Giving the outcome as required.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.