https://file.aiquickdraw.com/mathbot/file/17271903913282sq77fyi.jpg

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.

Solve the following right-angled triangles and trigonometric equations provided in the image.

Learn how to solve right-angled triangles and trigonometric equations using the Pythagorean theorem and basic trigonometric ratios. Problems include finding unknown sides and angles, applying trigonometric identities, and verifying triangle properties. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation