I would like to know how to calculate the angle of the 2 red lines based on screenshot below.
I have a circle and draw two lines of 45 degrees (black lines forming 90 degree angle). The red lines are the result of the 2 rectangles drawn on the circle. I would like to calculate the angle of the 2 red lines.
We don't know anything else except the length and angle as seen in the screenshot.
needed angle:
I don't know if I am on the right forum. If not please let me know, I will delete the question and go to the forum I supposed to ask this question on.
With the correct dimensions, following vectors will answer the question. If we go a distance $\delta$ along the $45^\circ$ line from the circle to the far edge of the rectangle, then we find the upper intersection $P$ of the rectangle with the circle by solving for the value of $\delta$ to make the vector \begin{align*} (1+\delta)\big(\frac1{\sqrt2},\frac1{\sqrt2}\big) + \frac12\big({-}\frac1{\sqrt2},\frac1{\sqrt2}\big) + \frac12\big({-}\frac1{\sqrt2},{-}\frac1{\sqrt2}\big)&=\frac1{\sqrt2}\big(1+\delta-1,1+\delta\big)\\ &= \frac1{\sqrt2}(1+\delta,\delta) \end{align*} lie on the circle. So its length must be $1$. This gives us the equation $$(1+\delta)^2+\delta^2 = 2,$$ so, by the quadratic formula, $\delta = \dfrac{\sqrt3-1}2$. That tells us that the point $P$ is $\dfrac1{\sqrt2}\big(\dfrac{\sqrt3+1}2,\dfrac{\sqrt3-1}2\big)$.
We can use dot product to find the angle $\theta$ between the vector $\overrightarrow{OP}$ and the $45^\circ$ line: $$\cos\theta = \frac{(1,1)}{\sqrt2}\cdot\frac{(\sqrt3+1,\sqrt3-1)}{2\sqrt2} = \frac{\sqrt3}2.$$ This means that $\theta = 30^\circ$. This means that half your red angle is $(90-30-45)^\circ = 15^\circ$ and the whole red angle is $30^\circ$. Quite amazing how that worked out!!
EDIT: I followed the vectorial approach because that's how I worked the problem with my original understanding of the dimensions. However, having arrived at the answer, I now realize there's an easy solution using just basic geometry. As pictured, below, we have rectangle $PQRS$ with $PS=QR=1$. Let $O$ be the center of the unit circle. Therefore, $\triangle OPS$ is equilateral (since the circle has radius $1$). Bisecting $\angle SOP$ gives the $30^\circ$ angle between the radius perpendicular to $\overline{QR}$ and $\overline{OP}$, as required.