https://file.aiquickdraw.com/mathbot/file/17254316086009s3rsb31.jpg

The image provides a mathematical problem involving a diagram with triangles and angles. Here is a summary of the given information:

### Given:
- \( AC = 16 \, \text{cm} \)
- \( CE = 12 \, \text{cm} \)
- \( \angle ABC = 90^\circ \)
- \( \angle ACE = 90^\circ \)
- \( \angle BAC = 6^\circ \)
- Line \( BD \) is a straight line.

### Part (a)
We need to explain why \( BD = 16 \sin \theta + 12 \cos \theta \).

### Part (b)
\( BD \) can be expressed in the form \( R \sin(\theta + \alpha) \), where \( R > 0 \) and \( 0^\circ < \alpha < 90^\circ \). We are to find the values of \( R \) and \( \alpha \).

---

Let's solve part (a) and part (b).

### Solution for Part (a):

Given the expression for \( BD = 16 \sin \theta + 12 \cos \theta \), this can be derived using the coordinates of points in a right triangle setup.

Let's assume:
- \( AC \) and \( CE \) represent the sides of right triangles with the given angles.
- Using trigonometry, we represent the horizontal and vertical components of \( BD \) using \( \sin \) and \( \cos \) functions.

For a point on line \( BD \):
- The component along \( AC \) contributes \( 16 \sin \theta \).
- The component along \( CE \) contributes \( 12 \cos \theta \).

Combining these, we get:
\[
BD = 16 \sin \theta + 12 \cos \theta.
\]

### Solution for Part (b):

To express \( BD \) in the form \( R \sin(\theta + \alpha) \), we use the sine addition formula:

\[
R \sin(\theta + \alpha) = R (\sin \theta \cos \alpha + \cos \theta \sin \alpha).
\]

Comparing coefficients, we equate:
\[
R \cos \alpha = 16, \quad R \sin \alpha = 12.
\]

Solving for \( R \):
\[
R = \sqrt{(16)^2 + (12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20.
\]

Solving for \( \alpha \):
\[
\cos \alpha = \frac{16}{20} = 0.8, \quad \sin \alpha = \frac{12}{20} = 0.6.
\]

Thus, \( \alpha = \arctan\left(\frac{12}{16}\right) = \arctan(0.75) \approx 36.87^\circ \).

### Final Answers:
- **(a)** The expression \( BD = 16 \sin \theta + 12 \cos \theta \) comes from the horizontal and vertical components.
- **(b)** \( R = 20 \), \( \alpha \approx 36.87^\circ \).

Would you like more details, or do you have any questions?

### Additional Related Questions:
1. How do we find the trigonometric components for different sides of a triangle?
2. What is the geometric significance of expressing \( BD \) in the form \( R \sin(\theta + \alpha) \)?
3. Can this method be generalized to other problems involving similar right triangles?
4. How does the choice of angles affect the expression of \( BD \)?
5. What are some practical applications of such trigonometric expressions in physics or engineering?

**Tip:** Remember, when dealing with trigonometric identities, it's often helpful to draw a diagram to visualize the angles and sides involved.

In the diagram, it is given that AC = 16 cm, CE = 12 cm, angle ABC = 90°, angle ACE = 90°, angle BAC = 6°, and BD is a straight line. a) Explain clearly why BD = 16sinθ + 12cosθ. b) BD can be expressed in the form Rsin(θ + α), where R > 0 and 0° < α < 90°. Find the values of R and α.

Explore the trigonometric problem involving a right triangle where BD is derived as 16sinθ + 12cosθ. Learn how to express BD in the form Rsin(θ + α) and calculate the values of R and α. Suitable for Grades 10-12 students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation