Given theta = (9)/(5), find the value of theta.

Substitute the given value $\sin \theta = \frac{9}{5}$ into the identity.

$$\left(\frac{9}{5}\right)^2 + \cos^2 \theta = 1$$

$$\cos^2 \theta = 1 - \frac{81}{25}$$

To subtract, express 1 as $\frac{25}{25}$:

$$\cos^2 \theta = \frac{25}{25} - \frac{81}{25}$$ $$\cos^2 \theta = \frac{25 - 81}{25}$$ $$\cos^2 \theta = -\frac{56}{25}$$

The square of any real number, including $\cos \theta$, cannot be negative. Since $\cos^2 \theta = -\frac{56}{25}$ is a negative value, there is no real value for $\cos \theta$ that satisfies this equation.

Additionally, the value $\sin \theta = \frac{9}{5}$ (which is 1.8) is outside the valid range for the sine function. The sine of any real angle must be between $-1$ and 1 (inclusive), i.e., $-1 \le \sin \theta \le 1$.
Since $\frac{9}{5} > 1$, there is no real angle $\theta$ for which $\sin \theta = \frac{9}{5}$. Therefore, no such acute angle exists.

Given the impossible condition, it is not possible to find a real value for $\cos \theta$.