Stuck on trigonometry question

[gsf600y]

Completely stuck on this trigonometry question:


Added some labels.....

So, obviously side "b" is easy enough, 8.544.

And I know that
sin(B) / b = sin(A) /a = sin(C) / c

So for the B/b angle/side:
0.866 / 8.544 = 0.101


But, I'm totally stumped at what to do next.


I could make it into some right angled triangles:

But not sure how that helps me, as neither triagle has enough information to use the basic SOH-CAH-TOA rules.

Hope someone can give me some pointers, thanks

 

[gsf600y]

(x+2)^2 + (2x-3)^2 = 73

Expand, factorise and rearrange to get x.... unless I'm missing something. Don't need to do any sinA triangle stuff...?

I really should have said in my opening post, I'm not good at maths, not even at GCSE level. I just don't have the knowledge of how to do what you're suggesting :O
I figured my answer would need to involved trigonometry rules, but perhaps that's where I'm going wrong - it's just another area of basic maths I need to know.

 

[gsf600y]

Substitute into the cosine rule formula and sort it from there.

a^2 = b^2 + c^2 - 2bccosA

I did think about the cosine rule, but then figured I had too many things missing to make it work.

 

[gsf600y]

Just reading about factorising and rearranging now......

 

[gsf600y]

This a prototype for v2 of the OCUK 2FA?

That's me locked out for life then

 

[gsf600y]

Using @lemonkettaz cosine rule:
73 = (X+2)² + (2X - 3)² -2(X+2)(2X-3)Cos(60)
73 = X² + 4X + 4 + 4X² -12X +9 -2X² -4X +3X +6
73 = 3X² -9X +19
0 = 3X² -9X -54
0 = X² - 3X -18

x = 6

Very interesting, thanks!
Going to take a while to digest that, but I'm sure that'll keep me busy!

Thanks everyone.

EDIT: I see I've probably labelled my images incorrectly....I stuck the known length as "b", but I see you've started with 73 as "a". I guess you want to start with the known value on one side of the equation, and go from there?

 

[gsf600y]

The OP gave himself the chance of resolving the question when he added the green perpendicular line.

Thanks. Not sure what I'd do next to calculate the green line height though - on the right hand triangle, I've only got 1 side length, and the 90 degree angle....but I don't think you can use the 90' angle for the basic SOH-CAH-TOA rules?