Tension woes simple mass problem

[veitch]

I thought I had this stuff figured out, but my answer is wrong. Any tips?
Thanks

Homework Statement

A sign hangs precariously from your prof's office door. Calculate the magnitude of the tension in string 1 if theta-1 = 26.10degrees, theta-2 = 54.87degrees, and the mass of the sign is 6.50 kg.

http://img89.imageshack.us/img89/1089/prob05signly6.th.gif http://g.imageshack.us/thpix.php

Homework Equations

\sumF_x = T_1x - T_2x = 0
\sumF_y = T_1y + T_2y - F_g = 0
F_g = mg

The Attempt at a Solution

X components:

T_1x = T₁ Cos 26.1
T_2x = T₂ Cos 54.87

Y components:

T_1y = T₁ Sin 26.1
T_2y = T₂ Sin 54.87

\sumF_x = T₁Cos 26.1 - T₂Cos 54.87 = 0
So...
T₁Cos 26.1 = T₂Cos 54.87
T₁ = T₂(Cos 54.87/Cos 26.1)

and

\sumF = T₁Sin 26.1 + T₂ Sin 54.87 = (6.5 kg)(9.8 m/s²)

Then I tried plugging that T₁ in there...

T₂(Cos 54.87/Cos 26.1)Sin 26.1 + T₂Sin 54.87 = 24.7 N
T₂(0.6408)Sin 26.1 + T₂Sin 54.87 = 24.7 N
T₂(0.2819) + T₂(0.8178) = 24.7 N
T₂(0.2819 + 0.8178) = 24.7 N
T₂ = 22.5 N

I can't attempt any more solutions as the assignment was due 20 minutes ago (it's online)... but I did poorly on my midterm yesterday so I really need to get a better understand of the material before it's too late... any advice would be greatly appreciated!

 

[Doc Al]

\sumF = T₁Sin 26.1 + T₂ Sin 54.87 = (6.5 kg)(9.8 m/s²)

Then I tried plugging that T₁ in there...

T₂(Cos 54.87/Cos 26.1)Sin 26.1 + T₂Sin 54.87 = 24.7 N

Your method looks fine, just be careful with the arithmetic: 6.5*9.8 ≠ 24.7.