Assume 𝜃 lies in quadrant 3 and the terminal side of 𝜃 is perpendicular to the line


*y* = −15 *x* + 3

**Part 1:** Determine *sin(𝜃)*
**Part 2:** Determine *sec(𝜃)*
*sin(𝜃)* = 
 
*sec(𝜃)* = 


To solve this problem, we need to use the geometric relationships of the given line and the properties of angles in the third quadrant.

### Part 1: Determine *sin(𝜃)*

The terminal side of the angle \( \theta \) is **perpendicular** to the line \( y = -15x + 3 \). The slope of this line is -15. Since the terminal side is perpendicular to this line, the slope of the terminal side must be the negative reciprocal of -15, which is:

\[
\text{slope of terminal side} = \frac{1}{15}
\]

This gives us the rise over run for a point on the terminal side of the angle. Consider a right triangle formed by a point on the terminal side, the origin, and the projection onto the x-axis. The sides of the triangle will be:

- Opposite side = 1 (rise)
- Adjacent side = 15 (run)

We can now calculate the hypotenuse using the Pythagorean theorem:

\[
r = \sqrt{1^2 + 15^2} = \sqrt{1 + 225} = \sqrt{226}
\]

In quadrant 3, both the \( x \)-coordinate and the \( y \)-coordinate are negative. Therefore, the opposite side is -1 and the adjacent side is -15.

Now, we can find \( \sin(\theta) \):

\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{-1}{\sqrt{226}}
\]

So:

\[
\sin(\theta) = \frac{-1}{\sqrt{226}}
\]

### Part 2: Determine *sec(𝜃)*

Next, we find \( \sec(\theta) \). Since \( \sec(\theta) = \frac{1}{\cos(\theta)} \), we first need to find \( \cos(\theta) \).

\[
\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{-15}{\sqrt{226}}
\]

Now, using this, we can determine \( \sec(\theta) \):

\[
\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{\frac{-15}{\sqrt{226}}} = \frac{\sqrt{226}}{-15}
\]

Thus:

\[
\sec(\theta) = \frac{\sqrt{226}}{-15}
\]

### Final Answers:

\[
\sin(\theta) = \frac{-1}{\sqrt{226}}
\]
\[
\sec(\theta) = \frac{\sqrt{226}}{-15}
\]

Would you like further details or have any questions?

Here are 5 related questions to consider:
1. How would the signs of trigonometric functions change in other quadrants?
2. What is the reciprocal of each trigonometric function?
3. How do we determine the cotangent given this scenario?
4. Can you explain how to derive the cosine from the sine of an angle?
5. How do you simplify expressions like \( \frac{1}{\sqrt{226}} \)?

**Tip**: When solving trigonometric problems involving perpendicular lines, always start by finding the slopes and use Pythagorean relationships to handle the geometry.

Assume 𝜃 lies in quadrant 3 and the terminal side of 𝜃 is perpendicular to the line y = −15x + 3. Part 1: Determine sin(𝜃). Part 2: Determine sec(𝜃).

Learn how to find sin(𝜃) and sec(𝜃) for an angle in quadrant 3, where the terminal side is perpendicular to the line y = −15x + 3. This solution involves understanding trigonometric relationships, slopes, and the Pythagorean theorem. Suitable for high school students in Grades 10-12.

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