Triangle trig

In the diagram, triangle ABC is right-angled at C. Also, 2sinB = 3tanA. Determine the measure of angle A.




Its small. REAL small. Click to zoom.

Thanks

If a graphing calculator is allowed, you can simply graph 3sin(x) and 2tan(90-x), which intersect in only one place within 0<x<90, x=60. So B=60 degrees and A=90-60=30 degrees.

Otherwise, you have:
B=90-A
2sin(90-A)=3tan(A)
2cos(A)=3tan(B)
2cos(A)=3sin(A)/cos(A)
2cos²(A)=3sin(A)
2(1-sin²(A))=3sin(A)

From here, you can substitute u=sin(A), solve the quadratic equation for u, then take the sin^-1 for whatever answer works, and I'm guessing you'll get A=60 degrees.

alrite, some problems here:

2sin(90-A)=3tan(A)
2cos(A)=3tan(B)
-on the left side, how do you change sin to cos like that?

similarly....
2cos²(A)=3sin(A)
2(1-sin²(A))=3sin(A)
-the LS again, what did you do to get that?

3rd: what u?, when you say u=sinA?

and lastly, last line i think you mean A = 30

and those are the only problems i have.
thanks!

"on the left side, how do you change sin to cos like that?"->It's a trigonometric identity for sin(a-b)

"-the LS again, what did you do to get that?"

sin²(x) + cos²(x) = 1; identity
1-sin²(x)=cos²(x)

"what u?, when you say u=sinA?" 2(1-sin²(A))=3sin(A)
Literally substitute u=sin(A) into the equation and solve: 2(1-u²)=3u
Then, once you have u, you can find a because you know u=sin(A)

"and lastly, last line i think you mean A = 30" er, oops, yeah, I meant 30

ic now, thanks, bfrsoccer

2cos(A)=3tan(B)

[quote]

I think you meant tan(A).

Oops, yeah, sorry.