Math Help - find angle between lines

1. find angle between lines

find the acute angle between the lines:

x - 2y + 1 = 0
and
y = 5x - 4

2. Originally Posted by deej813
find the acute angle between the lines:

x - 2y + 1 = 0
and
y = 5x - 4
1. $x - 2y + 1 = 0 ~\implies~y=\frac12 x +\frac12$

$y = 5x - 4$

2. The angle between the first line and the horizontal is $\alpha$. Then you know that $\tan(\alpha) = \frac12$ .

The angle between the second line and the horizontal is $\beta$. Then you know that $\tan(\beta) = 5$ .

3. The angle which is included by the 2 lines is $(\beta - \alpha)$

4. Use the property:

$\tan(\beta - \alpha) = \dfrac{\tan(\beta) - \tan(\alpha)}{1+ \tan(\alpha) \cdot \tan(\beta)}$

5. Plug in the values you know. You should come out with: $(\beta - \alpha) \approx 52.125^\circ$

3. Completely forgot about that identity, thanks Good to know.

4. ok thanks
but what is that property that you used?
how do you get it?
how does it work?

5. Originally Posted by deej813
ok thanks
but what is that property that you used?
This formula belongs to a set of equations concerning the trigonometric functions of sums (or differences) of angles.
how do you get it?
I learned it nearly 50 years ago - so I don't get it but I have it
how does it work?
Fine!
As I've mentioned you only have to plug in the known values:

$\tan(\beta - \alpha) = \dfrac{5 - \frac12}{1+ 5 \cdot \frac12}= \dfrac{\frac92}{\frac72}=\dfrac97
$

Now you can calculate the angle using the $\arctan-$ or $tan^{-1}$ function.

6. ok then so its just a formula?
is there another way to do it seen as i havnt learnt that yet or is this it?

7. Originally Posted by deej813
ok then so its just a formula?
is there another way to do it seen as i havnt learnt that yet or is this it?
Of course there is.

Solve for $\alpha$: $\alpha = \tan^{-1} \frac{1}{2}$ (use a calculator).

Solve for $\beta$: $\beta = \tan^{-1} 5$ (use a calculator).

Calculate $\beta - \alpha$.

8. Originally Posted by deej813
what is that property that you used?
how do you get it?
You don't have to remember the formula for $\tan(\beta-\alpha)$, provided that you know the formulas for $\sin(\beta-\alpha)$ and $\cos(\beta-\alpha)$ (which are definitely things that are useful to remember). In fact $\tan(\beta-\alpha) = \frac{\sin(\beta-\alpha)}{\cos(\beta-\alpha)} = \frac {\sin\beta\cos\alpha-\cos\beta\sin\alpha} {\cos\beta\cos\alpha+\sin\beta\sin\alpha}$. Now divide top and bottom by $\cos\beta\cos\alpha$ and you get the result $\tan(\beta-\alpha) = \frac{\tan\beta-\tan\alpha}{1+\tan\beta\tan\alpha}$.