How do i solve for iii /b/? Also did I do the first 2 correctly?

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

>>723463059
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?

>>723463059
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.

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763KB, 1379x723px

>>723464540

>>723464540
Why can he assume angle D is a right angle?

>>723464163
Wait no you're right. Cosine rule. I'm a faggot.

First one is right too.

So refer to>>723464225

>>723464708
He doesn't need to. He creates his own right angle by dropping a line between angle D and line EF.

>>723463059
sin30 lol, dont miss these: °

>>723464772
Right, but he still doesn't know the angles in DGF

>>723464843
Irrelevant

>>723464540
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.

>>723464918
He knows two of the sides.

>>723465005
Sorry-triangle DFG.

>>723465002
4.867258771

>>723463059
>>>/rules/global/2

>>723465042
You're right, my bad