https://i.imgur.com/isLpH5w.png
Not entirely sure if its correct. Need help...x = [20:6:68];
t1=atan((8-x.*tan(8*pi/180))/x);
t=atan((22-x.*tan(8*pi/180))/x)-t1;
Table = [t1' t'];
disp(Table)
https://i.imgur.com/isLpH5w.png
Not entirely sure if its correct. Need help...x = [20:6:68];
t1=atan((8-x.*tan(8*pi/180))/x);
t=atan((22-x.*tan(8*pi/180))/x)-t1;
Table = [t1' t'];
disp(Table)
Let the positive x-axis be to the left.
The top of the screen is at $p_{t}=(0,30)$
The bottom of the screen at $p_{b}=(0,8)$
A point on the sloped line at distance $x$ from the screen is at $p=(x, x\tan(8^\circ))$
A vector from the point $p$ to the top of the screen is $v_t=p_t-p=(-x,30-x\tan(8^\circ))$
A vector from the point $p$ to the bottom of the screen is $v_b=p_b - p = (-x, 8-x\tan(8^\circ))$
The angle between these two vectors is $\theta = \arccos\left(\dfrac{v_t \cdot v_b}{|v_t||v_b|}\right)$
This produces different results than your formula.
Rewrite your program to perform the above.