Let's begin by solving the problems listed in the image.

Part 1: Trigonometric Equations

i. $2\sin^2\theta = \frac{1}{2}$

To solve for $\theta$:

1. Divide both sides by 2: $\sin^2\theta = \frac{1}{4}$
2. Take the square root of both sides: $\sin\theta = \pm \frac{1}{2}$
3. Solving for $\theta$: $\theta = 30^\circ, 150^\circ \quad (\text{for } \sin\theta = \frac{1}{2})$ $\theta = 210^\circ, 330^\circ \quad (\text{for } \sin\theta = -\frac{1}{2})$

ii. $\sin\theta = \cos\theta$

We know that: $\sin\theta = \cos\theta \implies \tan\theta = 1$ Thus, $\theta = 45^\circ$, and for general solutions: $\theta = 45^\circ + 180^\circ n, \quad n \in \mathbb{Z}$

Part 2: Solving Right-Angled Triangles

Now, let's proceed with solving the right-angled triangles.

i. $\Delta ABC$, $\angle C = 90^\circ$, $a = 4 \, \text{cm}, \, c = 4\sqrt{2} \, \text{cm}$

We will use the Pythagorean theorem to find side $b$: $c^2 = a^2 + b^2$ $(4\sqrt{2})^2 = 4^2 + b^2 \quad \implies \quad 32 = 16 + b^2 \quad \implies \quad b^2 = 16 \quad \implies \quad b = 4 \, \text{cm}$

ii. $\Delta PQR$, $\angle Q = 90^\circ, \, \angle P = 60^\circ, \, PQ = 4\sqrt{3} \, \text{cm}$

Since we have one angle and one side, we can use trigonometric ratios to solve for the other sides: $\sin 60^\circ = \frac{QR}{PQ} \quad \implies \quad QR = PQ \times \sin 60^\circ = 4\sqrt{3} \times \frac{\sqrt{3}}{2} = 6 \, \text{cm}$ $\cos 60^\circ = \frac{PR}{PQ} \quad \implies \quad PR = PQ \times \cos 60^\circ = 4\sqrt{3} \times \frac{1}{2} = 2\sqrt{3} \, \text{cm}$

iii. $\Delta PQR$, $\angle R = 90^\circ, \, PR = 8 \, \text{m}, \, RQ = 8 \, \text{m}$

This is an isosceles right triangle, so $PQ$ can be calculated using: $PQ = PR \times \sqrt{2} = 8\sqrt{2} \, \text{m}$

iv. $\Delta XYZ$, $\angle X = 90^\circ, \, YZ = 16 \, \text{m}, \, XZ = 8 \, \text{m}$

Using the Pythagorean theorem: $YZ^2 = XY^2 + XZ^2 \quad \implies \quad 16^2 = XY^2 + 8^2 \quad \implies \quad 256 = XY^2 + 64 \quad \implies \quad XY^2 = 192$ $XY = \sqrt{192} = 8\sqrt{3} \, \text{m}$

v. $\Delta LMN$, $LM = 10 \, \text{cm}, \, MN = 8 \, \text{cm}, \, NL = 6 \, \text{cm}$

Since all three sides are given, we can verify if it's a right triangle by checking the Pythagorean theorem: $LM^2 = MN^2 + NL^2 \quad \implies \quad 10^2 = 8^2 + 6^2 \quad \implies \quad 100 = 64 + 36$ This is a valid right triangle.

Would you like further details on any of these solutions, or have any questions about the process?

Here are 5 related questions for further exploration:

1. How can we generalize solutions for trigonometric equations like $\sin^2 \theta = k$?
2. What is the method to find the area of right-angled triangles?
3. How can we apply the sine and cosine rules to non-right-angled triangles?
4. What are the steps for solving triangles using the tangent function?
5. How does the Pythagorean theorem extend to 3D geometry?

Tip: When solving trigonometric equations, always check for all possible solutions within the given range of angles.