More Complex Trig eqn, is it possible? (S*(tan(2b-a))) + D = S*(sin(a)) - (4/S)Sin(b)

[gbostock01]

Been working on this for a while now and can't solve it, was wondering if it's possible and how I'd then solve it?!
I know all the values except a.
Solve for a:

(S*(tan(2b-a))) + D = S*(sin(a)) - (4/S)Sin(b)



Alternatively here are are some values inserted.
(40*(tan(73.74-a))) + 30 = 40*(sin(a)) - (3/50)


Many thanks
George

 

[Deleted member 4993]

Been working on this for a while now and can't solve it, was wondering if it's possible and how I'd then solve it?!
I know all the values except a.
Solve for a:

(S*(tan(2b-a))) + D = S*(sin(a)) - (4/S)Sin(b)



Alternatively here are are some values inserted.
(40*(tan(73.74-a))) + 30 = 40*(sin(a)) - (3/50) .... where did Sin(b) go??


Many thanks
George

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[Ishuda]

Been working on this for a while now and can't solve it, was wondering if it's possible and how I'd then solve it?!
I know all the values except a.
Solve for a:

(S*(tan(2b-a))) + D = S*(sin(a)) - (4/S)Sin(b)



Alternatively here are are some values inserted.
(40*(tan(73.74-a))) + 30 = 40*(sin(a)) - (3/50)


Many thanks
George

As a hint, if you care to take it, I believe that this can be turned into a quartic equation in tan(2b-a), i.e.
x⁴ + A x³ + B x² + C x + D = 0
where x=tan(2b-a).