1. **Problem:** Solve triangle \(\triangle ABC\) given \(A=36^\circ\) and \(c=9\). We need to find sides \(a, b\) and angles \(B, C\).\n\n2. **Formula:** Use Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ and angle sum property: $$A + B + C = 180^\circ$$.\n\n3. **Step 1:** Find angle \(C\) using Law of Sines if possible or find \(B\) first. But we only have \(A\) and \(c\), so we need more info or assume \(b\) or \(a\) is unknown. Since only \(A\) and \(c\) are given, we cannot solve uniquely without more data.\n\n**Note:** The problem likely expects to use Law of Sines with given \(A=36^\circ\) and \(c=9\) to find other parts if more info is given, but here only one side and one angle are given, so the triangle is not solvable uniquely.\n\n7. **Problem:** Convert \(\frac{9\pi}{4}\) radians to degrees.\n\n**Formula:** $$\text{Degrees} = \text{Radians} \times \frac{180^\circ}{\pi}$$\n\n**Calculation:** $$\frac{9\pi}{4} \times \frac{180^\circ}{\pi} = \frac{9 \times 180^\circ}{4} = 405^\circ$$\n\n**Answer:** \(\frac{9\pi}{4} = 405^\circ\)\n\n10. **Problem:** Find \(\cos(-90^\circ)\).\n\n**Rule:** Cosine is an even function, so \(\cos(-\theta) = \cos \theta\).\n\n**Calculation:** \(\cos(-90^\circ) = \cos 90^\circ = 0\)\n\n**Answer:** 0\n\n13. **Problem:** Find \(\sec\left(-\frac{9\pi}{4}\right)\).\n\n**Step 1:** Recall \(\sec \theta = \frac{1}{\cos \theta}\).\n\n**Step 2:** Simplify angle \(-\frac{9\pi}{4}\) by adding \(2\pi\) multiples:\n$$-\frac{9\pi}{4} + 4\pi = -\frac{9\pi}{4} + \frac{16\pi}{4} = \frac{7\pi}{4}$$\n\n**Step 3:** Find \(\cos \frac{7\pi}{4}\). Since \(\frac{7\pi}{4} = 315^\circ\), \(\cos 315^\circ = \frac{\sqrt{2}}{2}\).\n\n**Step 4:** Calculate \(\sec \left(-\frac{9\pi}{4}\right) = \frac{1}{\cos \frac{7\pi}{4}} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}\).\n\n**Answer:** \(\sqrt{2}\)\n\n16. **Problem:** Given point \(P\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) on the unit circle, find \(\cos \theta\) and \(\sin \theta\).\n\n**Rule:** On the unit circle, coordinates correspond to \((\cos \theta, \sin \theta)\).\n\n**Answer:** \(\cos \theta = \frac{1}{2}\), \(\sin \theta = \frac{\sqrt{3}}{2}\)\n\n17. **Problem:** Which angle has tangent and sine both negative? Options: 65°, 120°, 265°, 310°\n\n**Step 1:** Recall signs of sine and tangent in quadrants:\n- Quadrant I: sin +, tan +\n- Quadrant II: sin +, tan -\n- Quadrant III: sin -, tan +\n- Quadrant IV: sin -, tan -\n\n**Step 2:** Check each angle's quadrant:\n- 65°: QI (sin +, tan +)\n- 120°: QII (sin +, tan -)\n- 265°: QIII (sin -, tan +)\n- 310°: QIV (sin -, tan -)\n\n**Answer:** 310° (Option D)\n\n22. **Problem:** Solve \(\triangle XYZ\) with \(y=15\), \(z=9\), and \(X=105^\circ\). Decide whether to use Law of Sines or Law of Cosines.\n\n**Step 1:** Since we have two sides and the included angle, use Law of Cosines to find side \(x\):\n$$x^2 = y^2 + z^2 - 2yz \cos X$$\n$$x^2 = 15^2 + 9^2 - 2 \times 15 \times 9 \times \cos 105^\circ$$\n\n**Step 2:** Calculate \(\cos 105^\circ \approx -0.2588\)\n$$x^2 = 225 + 81 - 2 \times 15 \times 9 \times (-0.2588) = 306 + 69.984 = 375.984$$\n$$x = \sqrt{375.984} \approx 19.4$$\n\n**Step 3:** Use Law of Sines to find angle \(Y\):\n$$\frac{\sin Y}{y} = \frac{\sin X}{x} \Rightarrow \sin Y = \frac{y \sin X}{x} = \frac{15 \times \sin 105^\circ}{19.4}$$\n\nCalculate \(\sin 105^\circ \approx 0.9659\):\n$$\sin Y = \frac{15 \times 0.9659}{19.4} = \frac{14.4885}{19.4} \approx 0.7471$$\n$$Y = \arcsin(0.7471) \approx 48.3^\circ$$\n\n**Step 4:** Find angle \(Z\):\n$$Z = 180^\circ - X - Y = 180^\circ - 105^\circ - 48.3^\circ = 26.7^\circ$$\n\n**Answer:** \(x \approx 19.4\), \(Y \approx 48^\circ\), \(Z \approx 27^\circ\) (rounded)