Finding the angle between two line equations

I need to find the angle between two lines in $y = mx + b$ form and they are:

\begin{equation*} y = 4x + 2~\text{and}~y = -x + 3 \end{equation*}

I have no idea how to solve this and if you could please consider that I'm in Grade 12 Calculus and Vectors. So my knowledge consist of adding vectors, derivatives, etc.

Somehow the answer is suppose to be 59 degrees but I don't know how to get that

Solution 1:

1. Find the slope of each line.
2. Find the angle of inclination of each line, using $\theta=\tan^{-1}m$. (Here, $\theta$ is the angle of inclination, $m$ is the slope.)
3. Subtract the two angles.
4. Handle the case where this difference is not an acute angle. (If you get a negative angle, take its absolute value. Also, when two lines intersect, they form two pairs of equal angles. Unless the lines are perpendicular, one pair will be acute and the other obtuse. You want to find the acute pair, so if you calculated the obtuse pair just subtract the value from pi radians or $180°$ to get the acute values.)

There are other ways, such as finding vectors on the lines and using their dot products. But the way I showed uses simple, basic trig methods.

The final answer is

$$\pi-\left|\tan^{-1}(4)-\tan^{-1}(-1)\right|$$

converted to degrees, if the calculator is in radians mode, or

$$180°-\left|\tan^{-1}(4)-\tan^{-1}(-1)\right|$$

if the calculator is in degrees mode.

Solution 2:

You should be knowing tan and arctan...

Angle between two lines of slope $m_1$ and $m_2$ is $ \tan^{-1}\dfrac{m_1 - m_2}{1+m_1 m_2} $

This is derived on basis of tangent of angular difference of given lines.

In your case $ m_1 = 4 $ and $ m_2 =-1$.

One angle the first line makes to x-axis $ \phi_1$ is in the first and the other $\phi_2$ is in the second quadrants, so you can take the difference angle in anticlockwise direction. Angle between them is $ 45^{\circ} +75.96^{\circ}=$

$ \tan^{-1}\dfrac{4 - (-1)}{1+(4 *-1)}=\tan^{-1}\dfrac{5}{-3}\approx 120.96^{\circ} $

all anti clockwise direction as shown in Geogebra construction.

Solution 3:

the first vector that parallel to $y=4x+2$ $$v_1=i+4j$$ the second vector is $$v_2=-i+j$$

then use the following formula $$\theta =\cos^{-1}(\frac{v_1.v_2}{|v_1||v_2|})$$ $$\theta =\cos^{-1}(\frac{(1\times-1+4\times1)}{\sqrt{17}\sqrt{2}})=59.03624347$$