How do i solve for iii Sup Forums? Also did I do the first 2 correctly?

How do i solve for iii Sup Forums? Also did I do the first 2 correctly?

No

That second one is like DF = atan( (3x-1)/(2x+1) ) or something isn't it. That's not right but it's closer.

Last one you solve 5(i)=(ii)

So once you need those solved first

You can't assume angle D is 90 degrees dumbass

Cos(60)=DE/EF

I don't think you have a bright future in maths ahead of you fam

Have you tried setting it to wumbo?

Make a line from D to line EF, where the angle formed at the intersection of the two lines is 90 degrees—call the intersection point G. You've just made a 30/60/90 triangle DEG. Solve for EG, subtract EG from EF, and you have another right triangle, DGF, and a known value for GF.

...

Why can he assume angle D is a right angle?

Wait no you're right. Cosine rule. I'm a faggot.

First one is right too.

So refer to

He doesn't need to. He creates his own right angle by dropping a line between angle D and line EF.

sin30 lol, dont miss these: °

Right, but he still doesn't know the angles in DGF

Irrelevant

Then you can find the length of GF using what you know about triangle DEG, and you wind up with values for two sides of the right triangle EFG, which is enough info to find the length of the third.

Yes, you can use the cosine rule, but this is how you do it geometrically.

He knows two of the sides.

Sorry-triangle DFG.

4.867258771

>>>/rules/global/2

You're right, my bad