1. **Problem statement:** Find the sine, cosine, and tangent of angle $D$ in a right triangle with sides opposite $D$ as 5, adjacent to $D$ as 5, and hypotenuse as $5\sqrt{3}$.\n\n2. **Recall definitions:** - $\sin(D) = \frac{\text{opposite}}{\text{hypotenuse}}$ - $\cos(D) = \frac{\text{adjacent}}{\text{hypotenuse}}$ - $\tan(D) = \frac{\text{opposite}}{\text{adjacent}}$ \n3. **Calculate sine:** $$\sin(D) = \frac{5}{5\sqrt{3}}$$ Simplify by canceling 5: $$\sin(D) = \frac{\cancel{5}}{\cancel{5}\sqrt{3}} = \frac{1}{\sqrt{3}}$$ Rationalize the denominator: $$\sin(D) = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$ \n4. **Calculate cosine:** $$\cos(D) = \frac{5}{5\sqrt{3}}$$ Simplify by canceling 5: $$\cos(D) = \frac{\cancel{5}}{\cancel{5}\sqrt{3}} = \frac{1}{\sqrt{3}}$$ Rationalize the denominator: $$\cos(D) = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$ \n5. **Calculate tangent:** $$\tan(D) = \frac{5}{5}$$ Simplify by canceling 5: $$\tan(D) = \frac{\cancel{5}}{\cancel{5}} = 1$$ \n**Final answers:** $$\sin(D) = \frac{\sqrt{3}}{3}, \quad \cos(D) = \frac{\sqrt{3}}{3}, \quad \tan(D) = 1$$