Is Δθ in a circle equal to the angle in a velocity vectors triangle?

[Hardikph]

How is Δθ in circle equals to angle in velocity vectors triangle?
I tried using simple geometry but I can't.

 

[Doc Al]

Try this:

What if r' were right on top of r? What is the angle between them? What is the angle between the velocity vectors?

Now move r' just a bit, an angle Δθ from r. How does the angle between the velocity vectors change?

 

[Clara Chung]

photo : http://s613.photobucket.com/user/Yan_Wa_Chung/media/Untitled_zpsqfu6lfi8.png.html
you have to consider angle a and b in the figure on the right. And with some 90 degree usage:)

 

[Clara Chung]

photo : http://s613.photobucket.com/user/Yan_Wa_Chung/media/Untitled_zpsqfu6lfi8.png.html
you have to consider angle a and b in the figure on the left. And with some 90 degree usage:)

That is not Vb in the figure on the right, that is -Vb(my mistake), so that dV = Va + (-Vb) ( vectors )

 

[Delta2]

Its because they are angles that have sides vertical to each other. r' is vertical to v' and r is vertical to v so the angle of (r',r) equals the angle of (v',v).

 

[Hardikph]

Delta² and Clara I appreciate your explanation but can you please describe some '90 degree usage'.

 

[Doc Al]

Another way to look at it: Realize that the velocity vectors are always 90° rotated compared to the r vectors. So if the angle that an r vector makes with the x-axis is θ, then the corresponding velocity vector must make an angle of θ + 90°.

 

[Clara Chung]

Delta² and Clara I appreciate your explanation but can you please describe some '90 degree usage'.

In the photo, θ+a = 90..(1) , a+b=90...(2), (1)-(2), θ=b , so it proves your statement "Δθ in circle equals to angle in velocity vectors triangle".
I hope it helps.

 

[James R]

View attachment 102908

One other point: the vector diagram on the right does not correspond to the vectors shown on the left.
Recall that ##\Delta \vec{v} = \vec{v}'-\vec{v}##. You also seem to have swapped the ##\vec{v}'## and the ##\vec{v}## in the diagram.