I am self studying and my exam is coming up in few days. I don't have a teacher to ask, so I decided to try Math.SE.
Why isn't the blue angle $30$? The weight force should always be perpendicular to a horizontal ground and the pink angle is $60$, so $90 - 60 = 30$. However when I use $\arctan(\frac{2}{3})$ it gives me 33.7 not 30.
Note that triangle is right angled triangle.
On
You are correct that the blue angle should be 30 if the purple one is 60. Nothing in the problem specifies the shape of the triangle fully-to my eye the third side looks about 4, not 5, so the leftmost angle is not known to be 90 degrees. In fact you can use the law of sines to find the leftmost angle: if $\theta$ is the angle at the top left of the vertical line, you have $\frac {\sin \theta}3=\frac {\sin 30^\circ}2=\frac 14, \theta = \arcsin \frac 34\approx 48.59^\circ$ and the leftmost angle of the triangle is $180-30-48.59=101.41$
You cannot do inverse tan of 2/3 to get that angle since that angle is not 90 degrees.