# Thread: Approximate change in area

1. ## Approximate change in area

Given the problem:

Two sides of a triangle measure 9 in. and 6 in. The measure of the inclusive angle changes from $35^\circ$ to $32^\circ$. Use differentials to approximate the change in the area of the triangle.

I used $A={\frac {1} {2}} ab \sin {C}$, where a and b are the lengths of the sides and C is the included angle, for my area function. I am just wondering if anyone has another approach for the area equation that would work well with this problem.

Thanks

2. Hello, McScruffy!

Your area formula is the best one to use.

Don't forget to convert all angles to radians.

3. Originally Posted by Soroban

Don't forget to convert all angles to radians.
Could you eplain why?

4. Originally Posted by VonNemo19
Could you eplain why?
Formulae such as $\frac{d}{d \theta} \cos \theta = - \sin \theta$ etc. are only valid when $\theta$ is measured in radians.

5. Originally Posted by mr fantastic
Formulae such as $\frac{d}{d \theta} \cos \theta = - \sin \theta$ etc. are only valid when $\theta$ is measured in radians.
I did not know that. I'm trying really hard to understand why. Don't tell me, give me a hint.

6. Originally Posted by VonNemo19
I did not know that. I'm trying really hard to understand why. Don't tell me, give me a hint.
Start from the first principles defination. The various trig limits (that you've probably taken for granted as standard forms) are found using radian measure. Google will turn up relevant pages. And there's an old thread somwehere that discusses the derivation of $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$ using a geometric approach ....

7. Originally Posted by mr fantastic
Start from the first principles defination. The various trig limits (that you've probably taken for granted as standard forms) are found using radian measure. Google will turn up relevant pages. And there's an old thread somwehere that discusses the derivation of $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$ using a geometric approach ....
Say no more. I'm very familiar with the proof. I now can see why the use radians was so covienient. It is because the area sector of the portion of the unit circle is squeezed between the two triangles as the angle tends to zero. Right? And using radians, the area sector is simply $\theta$. Is this right or wrong?

8. Originally Posted by VonNemo19
Say no more. I'm very familiar with the proof. I now can see why the use radians was so covienient. It is because the area sector of the portion of the unit circle is squeezed between the two triangles as the angle tends to zero. Right? And using radians, the area sector is simply $\theta$. Is this right or wrong?
Looks OK.