1. **State the problem:** Given a right triangle with sides $c=9$, $b=7$, and $a=130$, find $\sin(A)$, $\cos(A)$, and $\tan(A)$ where angle $C$ is the right angle. 2. **Check the given data:** Since $a=130$ is much larger than $b=7$ and $c=9$, this cannot be a right triangle with these side lengths because the hypotenuse should be the longest side. We assume $a$ is the hypotenuse opposite angle $A$. 3. **Find side $a$ if needed:** Normally, in a right triangle, $a^2 = b^2 + c^2$. Calculate: $$a = \sqrt{b^2 + c^2} = \sqrt{7^2 + 9^2} = \sqrt{49 + 81} = \sqrt{130} \approx 11.4$$ 4. **Since $a=130$ is given, it conflicts with the Pythagorean theorem.** We will proceed assuming $a = \sqrt{130}$ as the hypotenuse. 5. **Calculate $\sin(A)$:** Angle $A$ is opposite side $a$, but in the triangle, side $a$ is opposite angle $A$. Since $a$ is hypotenuse, $\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{b}{a} = \frac{7}{\sqrt{130}}$. 6. **Simplify $\sin(A)$:** $$\sin(A) = \frac{7}{\sqrt{130}} = \frac{7\sqrt{130}}{130}$$ 7. **Calculate $\cos(A)$:** $$\cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{c}{a} = \frac{9}{\sqrt{130}} = \frac{9\sqrt{130}}{130}$$ 8. **Calculate $\tan(A)$:** $$\tan(A) = \frac{\sin(A)}{\cos(A)} = \frac{\frac{7}{\sqrt{130}}}{\frac{9}{\sqrt{130}}} = \frac{7}{9}$$ **Final answers:** $$\sin(A) = \frac{7\sqrt{130}}{130} \approx 0.613$$ $$\cos(A) = \frac{9\sqrt{130}}{130} \approx 0.789$$ $$\tan(A) = \frac{7}{9} \approx 0.778$$