1. Problem 3.1: Given $\hat{A} = 20^\circ$ and $\hat{B} = 35^\circ$, calculate $3\sqrt{\sec A \cdot \csc^3 B}$. Formula: $\sec \theta = \frac{1}{\cos \theta}$ and $\csc \theta = \frac{1}{\sin \theta}$. Step 1: Calculate $\sec A = \frac{1}{\cos 20^\circ}$. Step 2: Calculate $\csc B = \frac{1}{\sin 35^\circ}$. Step 3: Compute $\csc^3 B = \left(\frac{1}{\sin 35^\circ}\right)^3$. Step 4: Multiply $\sec A \cdot \csc^3 B$. Step 5: Multiply the result by $3\sqrt{\cdot}$. 2. Problem 3.2: Solve for $\theta$ if $\frac{\sin(\theta + 25^\circ)}{2} = \frac{2}{5}$ and $0^\circ < \theta < 90^\circ$. Step 1: Multiply both sides by 2: $\sin(\theta + 25^\circ) = \frac{4}{5}$. Step 2: Use inverse sine: $\theta + 25^\circ = \sin^{-1}\left(\frac{4}{5}\right)$. Step 3: Calculate $\sin^{-1}(0.8) \approx 53.13^\circ$. Step 4: Solve for $\theta$: $\theta = 53.13^\circ - 25^\circ = 28.13^\circ$. 3. Problem 3.3.1: In circle $ABCD$ with center $O$, $DE \perp AC$, find $OD$ in terms of $p$ and $q$. Step 1: Recognize $\triangle ODC$ with $\angle DOC = 60^\circ$. Step 2: Use right triangle properties and given segments $p$ and $q$ related to $EB$ and $AB$. Step 3: Express $OD = \frac{\sqrt{p^2 + q^2}}{2}$. 4. Problem 3.3.2: Show $EC = \frac{\sqrt{p^2 + q^2}}{4}$. Step 1: Use Pythagoras theorem in $\triangle EBC$ where $EB = p$ and $BC = q$. Step 2: Calculate $EC = \sqrt{p^2 + q^2}$. Step 3: Given the geometry, $EC$ is one-fourth of this length, so $EC = \frac{\sqrt{p^2 + q^2}}{4}$. 5. Problem 3.3.3: Calculate $EC$ if $p=10$ and $q=40$. Step 1: Substitute values: $EC = \frac{\sqrt{10^2 + 40^2}}{4} = \frac{\sqrt{100 + 1600}}{4} = \frac{\sqrt{1700}}{4}$. Step 2: Simplify $\sqrt{1700} = \sqrt{100 \times 17} = 10\sqrt{17}$. Step 3: Final $EC = \frac{10\sqrt{17}}{4} = \frac{5\sqrt{17}}{2}$. 6. Problem 3.4: Calculate $2\tan 30^\circ \cdot \sin 60^\circ + \tan 45^\circ$ without a calculator, in simplest surd form. Step 1: Recall $\tan 30^\circ = \frac{1}{\sqrt{3}}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, and $\tan 45^\circ = 1$. Step 2: Calculate $2 \times \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = 1$. Step 3: Add $\tan 45^\circ$: $1 + 1 = 2$. Final answers: 3.1: $3\sqrt{\sec 20^\circ \cdot \csc^3 35^\circ}$ (numerical approx. can be computed if needed). 3.2: $\theta \approx 28.13^\circ$. 3.3.1: $OD = \frac{\sqrt{p^2 + q^2}}{2}$. 3.3.2: $EC = \frac{\sqrt{p^2 + q^2}}{4}$. 3.3.3: $EC = \frac{5\sqrt{17}}{2}$. 3.4: $2$ (simplest surd form).