1. ## line equations, trig

The lines l1 and l2, with equations y=2x and 3y=x−1 respectively, are drawn on the same set of axes. Given that the scales are the same on both axes and that the angles that l1 and l2 make with the positive x-axis are A and B respectively,

(a) write down the value of tan A and the value of tan B;

(b) without using your calculator, work out the acute angle between l1 and l2.

Really dont understand this question, how do I relate tanA and tanB with the equations?

2. Originally Posted by Tweety
Really dont understand this question, how do I relate tanA and tanB with the equations?
$\displaystyle \tan{A} = 2$

$\displaystyle \tan{B} = \frac{1}{3}$

$\displaystyle \tan(A-B) = \frac{\tan{A}-\tan{B}}{1 + \tan{A}\tan{B}}$

3. Originally Posted by skeeter
$\displaystyle \tan{A} = 2$

$\displaystyle \tan{B} = \frac{1}{3}$

$\displaystyle \tan(A-B) = \frac{\tan{A}-\tan{B}}{1 + \tan{A}\tan{B}}$
Thanks, I know what the answers are but don't know why ? Could you please explain,

Cause I thought of putting y=0 and x=0 and drawing the lines and still can't understand the answers. I can understand that to fine the acute angle, you would have to minus A from B.

4. Originally Posted by Tweety
Thanks, I know what the answers are but don't know why ? Could you please explain,

Cause I thought of putting y=0 and x=0 and drawing the lines and still can't understand the answers. I can understand that to fine the acute angle, you would have to minus A from B.
Hi Tweety,

The tangent ratio is the same as the slope. So that's why

$\displaystyle \tan A=2$ and $\displaystyle \tan B=\frac{1}{3}$

As far as the acute angle (let's call it $\displaystyle \theta$), formed by the two lines, you can use the exterior angle theorem to deduce that:

$\displaystyle \angle A - \angle B = \angle \theta$

Follow the previous post to determine that $\displaystyle \tan \theta = 1$. Thus $\displaystyle \angle \theta = 45^{\circ}$

5. Hi Tweety,

the slope of a line can be represented by placing a right-angled
triangle against it.

If the line has a positive slope, place the triangle under it,
so that the tangent of the lower left corner angle is opp/adj.
This is the exact same way to calculate the line slope,
using the differences between co-ordinates.

If the line has a negative slope, use the negative of the triangle's
tangent. So, a negative tangent means the slope of a negative-going line.

tan(angle) = line slope.