Trigonometric problem

[chandra21]

We assume angle POQ as del theta and apply sine on it.. are we take angle OPQ as right angle ...if so then how can OY is paroendicular to PQ again?

 

[Steven G]

The image is too small and not orientated correctly.

 

[chandra21]

Now?

 

[Dr.Peterson]

Here's a better image:

 

[Dr.Peterson]

The statement [MATH]r(r + \Delta r) \sin \Delta\theta = 2\Delta POQ[/MATH] uses the formula for area of a triangle, [MATH]Area = \frac{1}{2}a b \sin(C)[/MATH]. Are you familiar with this formula?

Nothing is assumed to be perpendicular, until they introduce OY as an altitude of the triangle to express the area a different way.

 

[chandra21]

I understand

 

[chandra21]

But when we divide both side by del t and take del t very small from where d theta/ dt comes from

 

[Dr.Peterson]

In the limit, [MATH]\frac{\Delta\theta}{\Delta t} \rightarrow \frac{d\theta}{d t}[/MATH]. Isn't that straightforward?