## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$\theta = 64.3^{\circ}$
We can find the area of the triangle: $A = \frac{\sqrt{3}}{4}~a^2$ $A = \frac{\sqrt{3}}{4}~(0.080~m)^2$ $A = 0.002771~m^2$ We can find the angle $\theta$: $\phi = B~A~cos~\theta$ $cos~\theta = \frac{\phi}{B~A}$ $\theta = cos^{-1}~(\frac{\phi}{B~A})$ $\theta = cos^{-1}~[~\frac{6.0\times 10^{-6}~Wb}{(5.0\times 10^{-3}~T)~(0.002771~m^2)}~]$ $\theta = 64.3^{\circ}$