Find the acute angle Show that both lines $r_1(t)=(3,9)+t(2,5)$ and $r_2(t)=(-5,6)+t(3,-1)$ contain the point $(1,4)$. Find the acute angle of intersection of these lines to the nearest degree.
I was able to prove that the point does cross both lines using parametric equations, but I do not know how to find the acute angle?
 A: i am assuming that the lines are $$r_1 = (3,9)+t(2,5), r_2 = (-5,6) + s(3, -1).$$ let the angle between the two lines be $\theta.$  then $$\cos \theta = \frac{(2,5) \cdot(3, -1)}{\sqrt{2^2 + 5^2}\sqrt{3^2 + (-1)^2}} =\frac{1}{\sqrt {290}} $$ this gives $$\theta = 86.63^\circ $$
A: Here is a method that uses trigonometry but not vector operations such as the dot product or cross product. Unfortunately, it does involve being careful of some special cases.
We can simply find the difference between the direction angles of the vectors that define the lines. The best way to find the direction angle is the $\operatorname{atan2}(y,x)$ function, which is similar to $\tan^{-1}(y/x)$ but for some values of $x$ and $y$ needs to be adjusted. Some programming environments have this function. If you don't have it, you can use $\tan^{-1}(y,x)$ perhaps with the needed adjustments. In your particular case, since $x>0$ for both vectors, i.e. both angles are in the first or fourth quadrants, no adjustment of the angle values is needed. Our calculation is therefore:
$$\theta=\left|\operatorname{atan2}(5,2)-\operatorname{atan2}(-1,3)\right|$$
$$=\left|\tan^{-1}\left(\frac 52\right)-\tan^{-1}\left(\frac {-1}3\right)\right|$$
$$=\left|68.1986^∘-(-18.4349^∘)\right|=86.6335^∘$$
In some cases you would need to add or subtract $360^∘$ but that was not needed here. (One such case would be the angle between $(-1,1)$ and $(-1,-1)$, which this method would give as $270^∘$ and should be adjusted to $-90^∘$ then to its absolute value.) Finally, the angle between two lines is given as an acute or right angle. If we had ended up with an angle between $90^∘$ and $180^∘$ we would have used the supplementary angle instead.
