Here's one for you

[jasoN-]

Don't know how to do this help appreciated

[Martoman]

Good Q.

The reason I choose to take tan of the angle i do is because the gradient makes that angle with the positive direction of the x axis.

Here's a solution... for...err...you.

[jasoN-]

okay I get it, thanks a lot

[kakar0t]

Hey Martoman, I understand the way you derived the magnitude for all the angles, but i don't get how you came to the conclusion that

m2 = tan(120+a)

Can you please elaborate why m2 equals that?

Cheers.

[brightsky]

Draw a vertical straight line from the point of intersection to the x-axis. Look at the triangle made up of the line with gradient , the x-axis and the vertical line we just drew.

Let the magnitude of our vertical straight line be and the magnitude of the x-axis that is part of our triangle be . Then . But we don't know what is at the moment.

To find the value of it, apply the trig knowledge you learnt. We know that . We know one angle (barring the right angle) of our triangle, and that angle is as Martoman found. Apply this rule:



But we also know that

So

[kakar0t]

Draw a vertical straight line from the point of intersection to the x-axis. Look at the triangle made up of the line with gradient , the x-axis and the vertical line we just drew.

I can't visualise this part Could you please elaborate on where the "vertical straight line" and the "point of intersection" are

This question = my downfall lol

[brightsky]

Draw a vertical straight line from the point of intersection to the x-axis. Look at the triangle made up of the line with gradient , the x-axis and the vertical line we just drew.

I can't visualise this part Could you please elaborate on where the "vertical straight line" and the "point of intersection" are

This question = my downfall lol

Oh woops, sorry, intersection is between the line with gradient and the line with gradient . Draw a vertical straight line from this point down to the x-axis. (I was drawing off Martoman's sketch btw).