Here's the solution for question 4: a) Step 1: Identify the known sides relative to angle $\alpha$. The side opposite to angle $\alpha$ is $AC = 2$. The hypotenuse is $AB = 4$. Step 2: Use the sine ratio to find angle $\alpha$. $$\sin(\alpha) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{2}{4} = \frac{1}{2}$$ Step 3: Calculate $\alpha$. $$\alpha = \arcsin\left(\frac{1}{2}\right)$$ $$\alpha = 30^\circ$$ Step 4: Use the sum of angles in a triangle to find angle $\theta$. The sum of angles in a triangle is $180^\circ$, and $\angle C = 90^\circ$. $$\theta + \alpha + 90^\circ = 180^\circ$$ $$\theta + 30^\circ + 90^\circ = 180^\circ$$ $$\theta + 120^\circ = 180^\circ$$ $$\theta = 180^\circ - 120^\circ$$ $$\theta = 60^\circ$$ The values of the angles are $\boxed{\alpha = 30^\circ}$ and $\boxed{\theta = 60^\circ}$. b) Step 1: Use the Pythagorean theorem to find the length of side $BC$. $$AC^2 + BC^2 = AB^2$$ Step 2: Substitute the known values $AC=2$ and $AB=4$. $$2^2 + BC^2 = 4^2$$ $$4 + BC^2 = 16$$ Step 3: Solve for $BC^2$. $$BC^2 = 16 - 4$$ $$BC^2 = 12$$ Step 4: Take the square root and simplify the surd. $$BC = \sqrt{12}$$ $$BC = \sqrt{4 \times 3}$$ $$BC = \sqrt{4} \times \sqrt{3}$$ $$BC = 2\sqrt{3}$$ The length of side $BC$ is $\boxed{2\sqrt{3}}$. That's 2 down. 3 left today — send the next one.