How may I calculate Earth's angular speed at specific point (The green dot for example)?
Note: I know that angular speed $= \dfrac{2\pi}{T}$ But how may I found $T$ in this case?
I found too that $R_{\text{new}}=R_{\text{old}}\sin (43)$ where $R_{old}$ stands for radius of Earth
How may I calculate earth's angular speed at specific point (The green dot for example)?
The angular velocity doesn't depend on the position of the green dot. Assuming uniform rotational speed it is simply:
$$\omega = \frac{2\pi}{T}$$
where $T$ is the period of rotation, in the case of Earth $T=24\mathrm{h}$.
The tangential velocity $v$ for a point is given by:
$$v=R\omega$$
where $R$ is the distance of the point to the axis of rotation (in your case $R_{earth}\sin\lambda$).
Notice, the angular speed $\omega$ of each and every on a rotating body remains constant irrespective of its position from the axis of rotation.
The angular speed at any point of a rotating body having period of one revolution $T$ (i.e. $24$ hrs for the Earth) is given as follows $$\omega=\frac{2\pi}{T}\ \ (\text{rad/s})$$ If $r$ is the normal distance of any specific point from the axis of rotation then its tangential velocity $v$ is given as $$v=r\omega \ \ (\text{m/s})$$
The normal distance of green dot from the axis of rotation is $R_{old}\sin\lambda$. Where $\lambda=43^\circ$. The angular speed $\omega$ of green dot will be constant while its tangential velocity $v$ will be $$v=(R_{old}\sin43^\circ)\omega$$
