Length of sides of a wire loop in a uniform magnetic field

[cestlavie]

Homework Statement

A wire loop with 60 turns is formed into a square with sides of length 𝑠. The loop is in the presence of a 1.00 T uniform magnetic field 𝐵 that points in the negative 𝑦 direction. The plane of the loop is tilted off the 𝑥-axis by $\theta=15$. If 𝑖=1.30 A of current flows through the loop and the loop experiences a torque of magnitude 0.0727 N⋅m , what are the lengths of the sides s of the square loop, in centimeters?

Relevant Equations

$A=x^2$
$\tau=NIABsin\theta$

If $\tau= 0.0727, N=60, i=1.3, B=1.0,$ and $\theta=15$, I tried the following calculation:
$\tau=NIABsin\theta$
$\tau=NIs^2Bsin\theta$
$s^2=\frac {\tau} {NIBsin\theta}=\frac {.0727} {60*1.3*1*sin(15)}=0.0632 m=6.32 cm$
The answer is probably right in front of me, but I don't know what I am doing wrong.

 

[Keith_McClary]

One thing, the dimensions of the last equation is length squared.

 

[cestlavie]

One thing, the dimensions of the last equation is length squared.

Oops! I forgot to add the square root! But my answer is for s, not s squared.
$s^2= \frac {\tau} {NIBsin\theta}$
$s=\sqrt {\frac {\tau} {NIBsin\theta}}=0.0632m = 6.32 cm$

 

[cestlavie]

I feel all kinds of stupid. I tried doing a different set of numbers and realized that my calculator was not in degrees. Thank you @Keith_McClary for your help.