https://file.aiquickdraw.com/mathbot/file/1729179489628db54bpap.jpg

Let's go through each triangle one by one, identify the relevant trigonometric ratio, and solve for the unknown side.

### a) Triangle \( \triangle ECD \)
- Given: Angle \( \angle E = 25^\circ \), adjacent side \( ED = 3.4 \, \text{m} \), and we are solving for the opposite side \( e = CD \).
- Trigonometric Ratio: Since we are dealing with the opposite and adjacent sides relative to the given angle, we use **tangent**:
  \[
  \tan(25^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{e}{3.4}
  \]
- Solving for \( e \):
  \[
  e = 3.4 \times \tan(25^\circ)
  \]
  Using a calculator:
  \[
  e \approx 3.4 \times 0.4663 = 1.5854 \, \text{m}
  \]
  So, \( e \approx 1.59 \, \text{m} \).

### b) Triangle \( \triangle PMQ \)
- Given: Angle \( \angle P = 13^\circ \), hypotenuse \( PQ = 103 \, \text{ft} \), and we are solving for the opposite side \( q = MQ \).
- Trigonometric Ratio: Since we have the opposite and hypotenuse, we use **sine**:
  \[
  \sin(13^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{q}{103}
  \]
- Solving for \( q \):
  \[
  q = 103 \times \sin(13^\circ)
  \]
  Using a calculator:
  \[
  q \approx 103 \times 0.2249 = 23.1667 \, \text{ft}
  \]
  So, \( q \approx 23.17 \, \text{ft} \).

### c) Triangle \( \triangle HGF \)
- Given: Angle \( \angle H = 43^\circ \), adjacent side \( HG = 43 \, \text{in} \), and we are solving for the hypotenuse \( g = HF \).
- Trigonometric Ratio: Since we are working with the adjacent and hypotenuse sides, we use **cosine**:
  \[
  \cos(43^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{43}{g}
  \]
- Solving for \( g \):
  \[
  g = \frac{43}{\cos(43^\circ)}
  \]
  Using a calculator:
  \[
  g \approx \frac{43}{0.7314} = 58.78 \, \text{in}
  \]
  So, \( g \approx 58.78 \, \text{in} \).

Let me know if you need further explanation or details on any step.

### Follow-up questions:
1. How do you decide which trigonometric function to use for each triangle?
2. Can we solve for all sides of a right triangle given just one side and an angle?
3. What is the difference between using sine, cosine, and tangent for right triangles?
4. How would you solve for an angle if given two sides of a triangle?
5. How can we apply these trigonometric ratios to solve real-world problems?

**Tip**: Always remember that for right triangles, the sine function relates the opposite side and hypotenuse, the cosine function relates the adjacent side and hypotenuse, and the tangent function relates the opposite and adjacent sides.

Identify the trigonometric ratio needed to find the unknown side length for each triangle, then solve for the unknown side length.

Learn how to solve for unknown side lengths in right triangles using trigonometric ratios. This problem explores the use of sine, cosine, and tangent functions to find the missing side lengths based on given angles and sides. Suitable for high school students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation